解：有个暴力是枚举两条链然后O（n）判定，可以得到15分。
还可以优化一下，枚举一条链，然后第二条链直接求两端点树上带权距离。可以做到O(m(n + m))，但是我用的树剖，跟上面那一档没啥区别Orz...
然后是链的部分，把所有链按照左端点小到大排序，左端点相同的时候按右端点小到大。
然后从右往左加。用一个树状数组维护前缀最大值。如果一个之前区间的左端点在我们当前区间的右端点内，就有交。
我们漏了一种情况：一个区间被另一个完全包含。于是新开一个树状数组维护即可。注意到这种情况在第一个树状数组中算出的答案不优，会被取代掉。
这样就有30分了。
1 /** 2 * There is no end though there is a start in space. ---Infinity. 3 * It has own power, it ruins, and it goes though there is a start also in the star. ---Finite. 4 * Only the person who was wisdom can read the most foolish one from the history. 5 * The fish that lives in the sea doesn't know the world in the land. 6 * It also ruins and goes if they have wisdom. 7 * It is funnier that man exceeds the speed of light than fish start living in the land. 8 * It can be said that this is an final ultimatum from the god to the people who can fight. 9 * 10 * Steins;Gate 11 */ 12 13 #include <bits/stdc++.h> 14 15 #define forson(x, i) for(int i = e[x]; i; i = edge[i].nex) 16 17 typedef long long LL; 18 const int N = 100010; 19 const LL INF = 9e18; 20 21 struct Edge { 22 int nex, v; 23 LL len; 24 }edge[N << 1]; int tp; 25 26 struct Node { 27 int x, y; 28 LL z, val; 29 inline bool operator < (const Node &w) const { 30 if(x != w.x) return x < w.x; 31 return y < w.y; 32 } 33 }node[N]; 34 35 int n, e[N], m, fa[N], deep[N]; 36 int num, pos[N], son[N], siz[N], top[N]; 37 LL Sum[N], sum[N << 2], d[N], tag[N << 2]; 38 39 inline void add(int x, int y, LL z) { 40 tp++; 41 edge[tp].v = y; 42 edge[tp].len = z; 43 edge[tp].nex = e[x]; 44 e[x] = tp; 45 return; 46 } 47 48 void DFS_1(int x, int f) { /// get fa son siz d deep 49 fa[x] = f; 50 deep[x] = deep[f] + 1; 51 siz[x] = 1; 52 forson(x, i) { 53 int y = edge[i].v; 54 if(y == f) continue; 55 d[y] = d[x] + edge[i].len; 56 DFS_1(y, x); 57 siz[x] += siz[y]; 58 if(siz[y] > siz[son[x]]) { 59 son[x] = y; 60 } 61 } 62 return; 63 } 64 65 void DFS_2(int x, int f) { /// get top id pos 66 pos[x] = ++num; 67 top[x] = f; 68 Sum[num] = d[x] - d[fa[x]]; 69 if(son[x]) { 70 DFS_2(son[x], f); 71 } 72 forson(x, i) { 73 int y = edge[i].v; 74 if(y == fa[x] || y == son[x]) continue; 75 DFS_2(y, y); 76 } 77 return; 78 } 79 80 inline int lca(int x, int y) { 81 while(top[x] != top[y]) { 82 if(deep[top[x]] > deep[top[y]]) { 83 x = fa[top[x]]; 84 } 85 else { 86 y = fa[top[y]]; 87 } 88 } 89 return deep[x] < deep[y] ? x : y; 90 } 91 92 inline LL dis(int x, int y) { 93 return d[x] + d[y] - 2 * d[lca(x, y)]; 94 } 95 96 inline void pushdown(int l, int r, int o) { 97 if(tag[o] != -1) { 98 if(tag[o] == 0) { 99 sum[o << 1] = sum[o << 1 | 1] = 0; 100 } 101 else { 102 int mid = (l + r) >> 1; 103 sum[o << 1] = Sum[mid] - Sum[l - 1]; 104 sum[o << 1 | 1] = Sum[r] - Sum[mid]; 105 } 106 tag[o << 1] = tag[o << 1 | 1] = tag[o]; 107 tag[o] = -1; 108 } 109 return; 110 } 111 112 void Add(int L, int R, int l, int r, int o) { 113 if(L <= l && r <= R) { 114 sum[o] = Sum[r] - Sum[l - 1]; 115 tag[o] = 1; 116 return; 117 } 118 int mid = (l + r) >> 1; 119 pushdown(l, r, o); 120 if(L <= mid) Add(L, R, l, mid, o << 1); 121 if(mid < R) Add(L, R, mid + 1, r, o << 1 | 1); 122 sum[o] = sum[o << 1] + sum[o << 1 | 1]; 123 return; 124 } 125 126 LL ask(int L, int R, int l, int r, int o) { 127 if(L <= l && r <= R) { 128 return sum[o]; 129 } 130 int mid = (l + r) >> 1; 131 pushdown(l, r, o); 132 LL ans = 0; 133 if(L <= mid) { 134 ans += ask(L, R, l, mid, o << 1); 135 } 136 if(mid < R) { 137 ans += ask(L, R, mid + 1, r, o << 1 | 1); 138 } 139 return ans; 140 } 141 142 inline void Add(int x, int y) { 143 while(top[x] != top[y]) { 144 if(deep[top[x]] > deep[top[y]]) { 145 std::swap(x, y); 146 } 147 Add(pos[top[y]], pos[y], 1, n, 1); 148 y = fa[top[y]]; 149 } 150 if(deep[x] > deep[y]) std::swap(x, y); 151 if(x != y) Add(pos[x] + 1, pos[y], 1, n, 1); 152 return; 153 } 154 155 inline LL ask(int x, int y) { 156 LL ans = 0; 157 while(top[x] != top[y]) { 158 if(deep[top[x]] > deep[top[y]]) { 159 std::swap(x, y); 160 } 161 ans += ask(pos[top[y]], pos[y], 1, n, 1); 162 y = fa[top[y]]; 163 } 164 if(deep[x] > deep[y]) std::swap(x, y); 165 if(x != y) ans += ask(pos[x] + 1, pos[y], 1, n, 1); 166 return ans; 167 } 168 169 inline void clear() { 170 memset(e + 1, 0, n * sizeof(int)); 171 memset(son + 1, 0, n * sizeof(int)); 172 tp = 0; 173 num = 0; 174 return; 175 } 176 /* 177 1 178 5 179 1 2 1 180 2 3 3 181 3 4 3 182 1 5 9 183 2 184 1 5 5 185 2 3 8 186 */ 187 188 namespace Line { 189 LL ta[N], ta2[N]; 190 inline LL ask(int i) { 191 LL ans = -INF; 192 for(; i; i -= i & (-i)) { 193 ans = std::max(ans, ta[i]); 194 } 195 return ans; 196 } 197 inline void Add(int i, LL v) { 198 for(; i <= n; i += i & (-i)) { 199 ta[i] = std::max(ta[i], v); 200 } 201 return; 202 } 203 inline LL ask2(int i) { 204 LL ans = -INF; 205 for(; i; i -= i & (-i)) { 206 ans = std::max(ans, ta2[i]); 207 } 208 return ans; 209 } 210 inline void Add2(int i, LL v) { 211 for(; i <= n; i += i & (-i)) { 212 ta2[i] = std::max(ta2[i], v); 213 } 214 return; 215 } 216 inline void solve() { 217 for(int i = 1; i <= n; i++) ta2[i] = ta[i] = -INF; 218 LL ans = -INF; 219 std::sort(node + 1, node + m + 1); 220 for(int i = m; i >= 1; i--) { 221 LL temp = ask(node[i].y - 1); 222 if(temp != -INF) { 223 ans = std::max(ans, temp + node[i].val - Sum[node[i].y]); 224 } 225 temp = ask2(node[i].y); 226 if(temp != -INF) { 227 ans = std::max(ans, temp + node[i].val); 228 } 229 Add(node[i].x, node[i].val + Sum[node[i].x]); 230 Add2(node[i].y, node[i].val + Sum[node[i].x] - Sum[node[i].y]); 231 } 232 if(ans == -INF) puts("F"); 233 else { 234 printf("%lld ", ans); 235 } 236 clear(); 237 return; 238 } 239 } 240 /* 241 50 242 2 243 1 2 914572059 244 3 245 1 2 213930211 246 1 2 0 247 2 2 0 248 */ 249 inline void solve() { 250 251 sum[1] = tag[1] = 0; 252 253 scanf("%d", &n); 254 LL z; 255 bool FLAG = 1; 256 for(int i = 1, x, y; i < n; i++) { 257 scanf("%d%d%lld", &x, &y, &z); 258 add(x, y, z); 259 add(y, x, z); 260 if(y != x + 1) { 261 FLAG = 0; 262 } 263 } 264 265 DFS_1(1, 0); 266 DFS_2(1, 1); 267 268 for(int i = 1; i <= n; i++) { 269 Sum[i] += Sum[i - 1]; 270 } 271 272 scanf("%d", &m); 273 for(int i = 1; i <= m; i++) { 274 scanf("%d%d%lld", &node[i].x, &node[i].y, &node[i].z); 275 if(node[i].x > node[i].y) std::swap(node[i].x, node[i].y); 276 else if(node[i].x == node[i].y) { 277 m--; 278 i--; 279 } 280 node[i].val = dis(node[i].x, node[i].y) - node[i].z; 281 //printf("node %d [%d %d] val = %lld ", i, node[i].x, node[i].y, node[i].val); 282 } 283 // solve 284 if(FLAG) { 285 Line::solve(); 286 return; 287 } 288 289 if(m > 300) { 290 clear(); 291 return; 292 } 293 294 LL ans = -INF; 295 for(int i = 1; i < m; i++) { 296 Add(node[i].x, node[i].y); 297 for(int j = i + 1; j <= m; j++) { 298 /// choose i j 299 LL Val = ask(node[j].x, node[j].y); 300 //printf("Val = %lld ", Val); 301 if(Val) { 302 ans = std::max(ans, node[i].val + node[j].val - Val); 303 } 304 } 305 tag[1] = sum[1] = 0; 306 } 307 if(ans == -INF) { 308 puts("F"); 309 } 310 else { 311 printf("%lld ", ans); 312 } 313 clear(); 314 return; 315 } 316 317 int main() { 318 319 int T; 320 scanf("%d", &T); 321 while(T--) solve(); 322 return 0; 323 }
接下来还有c = 0的部分分：枚举每条边，统计所有过它的链中权值最大的两条更新答案。拿线段树合并实现。
1 #include <bits/stdc++.h> 2 3 #define forson(x, i) for(int i = e[x]; i; i = edge[i].nex) 4 5 typedef long long LL; 6 const int N = 100010; 7 const LL INF = 9e18; 8 9 struct Edge { 10 int nex, v; 11 LL len; 12 }edge[N << 1]; int tp; 13 14 struct Node { 15 int x, y; 16 LL z, val; 17 inline bool operator < (const Node &w) const { 18 if(x != w.x) return x < w.x; 19 return y < w.y; 20 } 21 }node[N]; 22 23 int n, e[N], m, fa[N], deep[N]; 24 int num, pos[N], son[N], siz[N], top[N]; 25 LL Sum[N], sum[N << 2], d[N], tag[N << 2]; 26 27 inline void add(int x, int y, LL z) { 28 tp++; 29 edge[tp].v = y; 30 edge[tp].len = z; 31 edge[tp].nex = e[x]; 32 e[x] = tp; 33 return; 34 } 35 36 void DFS_1(int x, int f) { /// get fa son siz d deep 37 fa[x] = f; 38 deep[x] = deep[f] + 1; 39 siz[x] = 1; 40 forson(x, i) { 41 int y = edge[i].v; 42 if(y == f) continue; 43 d[y] = d[x] + edge[i].len; 44 DFS_1(y, x); 45 siz[x] += siz[y]; 46 if(siz[y] > siz[son[x]]) { 47 son[x] = y; 48 } 49 } 50 return; 51 } 52 53 void DFS_2(int x, int f) { /// get top id pos 54 pos[x] = ++num; 55 top[x] = f; 56 Sum[num] = d[x] - d[fa[x]]; 57 if(son[x]) { 58 DFS_2(son[x], f); 59 } 60 forson(x, i) { 61 int y = edge[i].v; 62 if(y == fa[x] || y == son[x]) continue; 63 DFS_2(y, y); 64 } 65 return; 66 } 67 68 inline int lca(int x, int y) { 69 while(top[x] != top[y]) { 70 if(deep[top[x]] > deep[top[y]]) { 71 x = fa[top[x]]; 72 } 73 else { 74 y = fa[top[y]]; 75 } 76 } 77 return deep[x] < deep[y] ? x : y; 78 } 79 80 inline LL dis(int x, int y) { 81 return d[x] + d[y] - 2 * d[lca(x, y)]; 82 } 83 84 inline void pushdown(int l, int r, int o) { 85 if(tag[o] != -1) { 86 if(tag[o] == 0) { 87 sum[o << 1] = sum[o << 1 | 1] = 0; 88 } 89 else { 90 int mid = (l + r) >> 1; 91 sum[o << 1] = Sum[mid] - Sum[l - 1]; 92 sum[o << 1 | 1] = Sum[r] - Sum[mid]; 93 } 94 tag[o << 1] = tag[o << 1 | 1] = tag[o]; 95 tag[o] = -1; 96 } 97 return; 98 } 99 100 void Add(int L, int R, int l, int r, int o) { 101 if(L <= l && r <= R) { 102 sum[o] = Sum[r] - Sum[l - 1]; 103 tag[o] = 1; 104 return; 105 } 106 int mid = (l + r) >> 1; 107 pushdown(l, r, o); 108 if(L <= mid) Add(L, R, l, mid, o << 1); 109 if(mid < R) Add(L, R, mid + 1, r, o << 1 | 1); 110 sum[o] = sum[o << 1] + sum[o << 1 | 1]; 111 return; 112 } 113 114 LL ask(int L, int R, int l, int r, int o) { 115 if(L <= l && r <= R) { 116 return sum[o]; 117 } 118 int mid = (l + r) >> 1; 119 pushdown(l, r, o); 120 LL ans = 0; 121 if(L <= mid) { 122 ans += ask(L, R, l, mid, o << 1); 123 } 124 if(mid < R) { 125 ans += ask(L, R, mid + 1, r, o << 1 | 1); 126 } 127 return ans; 128 } 129 130 inline void Add(int x, int y) { 131 while(top[x] != top[y]) { 132 if(deep[top[x]] > deep[top[y]]) { 133 std::swap(x, y); 134 } 135 Add(pos[top[y]], pos[y], 1, n, 1); 136 y = fa[top[y]]; 137 } 138 if(deep[x] > deep[y]) std::swap(x, y); 139 if(x != y) Add(pos[x] + 1, pos[y], 1, n, 1); 140 return; 141 } 142 143 inline LL ask(int x, int y) { 144 LL ans = 0; 145 while(top[x] != top[y]) { 146 if(deep[top[x]] > deep[top[y]]) { 147 std::swap(x, y); 148 } 149 ans += ask(pos[top[y]], pos[y], 1, n, 1); 150 y = fa[top[y]]; 151 } 152 if(deep[x] > deep[y]) std::swap(x, y); 153 if(x != y) ans += ask(pos[x] + 1, pos[y], 1, n, 1); 154 return ans; 155 } 156 157 inline void clear() { 158 memset(e + 1, 0, n * sizeof(int)); 159 memset(son + 1, 0, n * sizeof(int)); 160 tp = 0; 161 num = 0; 162 return; 163 } 164 /* 165 1 166 5 167 1 2 1 168 2 3 3 169 3 4 3 170 1 5 9 171 2 172 1 5 5 173 2 3 8 174 */ 175 176 namespace Line { 177 LL ta[N], ta2[N]; 178 inline LL ask(int i) { 179 LL ans = -INF; 180 for(; i; i -= i & (-i)) { 181 ans = std::max(ans, ta[i]); 182 } 183 return ans; 184 } 185 inline void Add(int i, LL v) { 186 for(; i <= n; i += i & (-i)) { 187 ta[i] = std::max(ta[i], v); 188 } 189 return; 190 } 191 inline LL ask2(int i) { 192 LL ans = -INF; 193 for(; i; i -= i & (-i)) { 194 ans = std::max(ans, ta2[i]); 195 } 196 return ans; 197 } 198 inline void Add2(int i, LL v) { 199 for(; i <= n; i += i & (-i)) { 200 ta2[i] = std::max(ta2[i], v); 201 } 202 return; 203 } 204 inline void solve() { 205 for(int i = 1; i <= n; i++) ta2[i] = ta[i] = -INF; 206 LL ans = -INF; 207 std::sort(node + 1, node + m + 1); 208 for(int i = m; i >= 1; i--) { 209 LL temp = ask(node[i].y - 1); 210 if(temp != -INF) { 211 ans = std::max(ans, temp + node[i].val - Sum[node[i].y]); 212 } 213 temp = ask2(node[i].y); 214 if(temp != -INF) { 215 ans = std::max(ans, temp + node[i].val); 216 } 217 Add(node[i].x, node[i].val + Sum[node[i].x]); 218 Add2(node[i].y, node[i].val + Sum[node[i].x] - Sum[node[i].y]); 219 } 220 if(ans == -INF) puts("F"); 221 else { 222 printf("%lld ", ans); 223 } 224 return; 225 } 226 } 227 228 namespace C0 { 229 int rt[N], xx; 230 LL Ans, X[N]; 231 std::vector<int> v[N]; 232 namespace seg { 233 const int M = 20000010; 234 int ls[M], rs[M], tot, cnt[M]; 235 LL sum[M]; 236 void insert(int p, int v, int l, int r, int &o) { 237 if(!o) { 238 o = ++tot; 239 sum[o] = cnt[o] = ls[o] = rs[o] = 0; 240 } 241 if(l == r) { 242 sum[o] += v * X[r]; 243 cnt[o] += v; 244 return; 245 } 246 int mid = (l + r) >> 1; 247 if(p <= mid) { 248 insert(p, v, l, mid, ls[o]); 249 } 250 else { 251 insert(p, v, mid + 1, r, rs[o]); 252 } 253 sum[o] = sum[ls[o]] + sum[rs[o]]; 254 cnt[o] = cnt[ls[o]] + cnt[rs[o]]; 255 return; 256 } 257 LL ask(int k, int l, int r, int o) { 258 if(l == r) { 259 return k * X[r]; 260 } 261 int mid = (l + r) >> 1; 262 if(k <= cnt[rs[o]]) { 263 return ask(k, mid + 1, r, rs[o]); 264 } 265 else { 266 return ask(k - cnt[rs[o]], l, mid, ls[o]) + sum[rs[o]]; 267 } 268 } 269 inline LL Ask(int x) { 270 if(cnt[rt[x]] < 2) return -INF; 271 return ask(2, 1, xx, rt[x]); 272 } 273 int merge(int x, int y) { 274 if(!x || !y) return x | y; 275 sum[x] += sum[y]; 276 cnt[x] += cnt[y]; 277 ls[x] = merge(ls[x], ls[y]); 278 rs[x] = merge(rs[x], rs[y]); 279 return x; 280 } 281 inline void Merge(int x, int y) { 282 rt[x] = merge(rt[x], rt[y]); 283 return; 284 } 285 } 286 void DFS(int x, int f) { 287 forson(x, i) { 288 int y = edge[i].v; 289 if(y == f) continue; 290 DFS(y, x); 291 Ans = std::max(Ans, seg::Ask(y)); 292 seg::Merge(x, y); 293 } 294 for(int i = 0; i < (int)v[x].size(); i++) { 295 int t = v[x][i]; 296 if(t > 0) { 297 seg::insert(t, 1, 1, xx, rt[x]); 298 } 299 else { 300 seg::insert(-t, -1, 1, xx, rt[x]); 301 } 302 } 303 return; 304 } 305 inline void solve() { 306 memset(rt + 1, 0, n * sizeof(int)); 307 seg::tot = 0; 308 Ans = -INF; 309 for(int i = 1; i <= n; i++) v[i].clear(); 310 for(int i = 1; i <= m; i++) { 311 X[i] = node[i].val; 312 } 313 std::sort(X + 1, X + m + 1); 314 xx = std::unique(X + 1, X + m + 1) - X - 1; 315 for(int i = 1; i <= m; i++) { 316 int val = std::lower_bound(X + 1, X + xx + 1, node[i].val) - X; 317 int x = node[i].x, y = node[i].y, z = lca(node[i].x, node[i].y); 318 v[x].push_back(val); 319 v[y].push_back(val); 320 v[z].push_back(-val); 321 v[z].push_back(-val); 322 } 323 DFS(1, 0); 324 if(Ans == -INF) puts("F"); 325 else printf("%lld ", Ans); 326 return; 327 } 328 } 329 330 /* 331 50 332 2 333 1 2 914572059 334 3 335 1 2 213930211 336 1 2 0 337 2 2 0 338 */ 339 inline void solve() { 340 341 sum[1] = tag[1] = 0; 342 343 scanf("%d", &n); 344 LL z; 345 bool FLAG = 1, c0 = 1; 346 for(int i = 1, x, y; i < n; i++) { 347 scanf("%d%d%lld", &x, &y, &z); 348 add(x, y, z); 349 add(y, x, z); 350 if(z) c0 = 0; 351 if(y != x + 1) { 352 FLAG = 0; 353 } 354 } 355 356 DFS_1(1, 0); 357 DFS_2(1, 1); 358 359 for(int i = 1; i <= n; i++) { 360 Sum[i] += Sum[i - 1]; 361 } 362 363 scanf("%d", &m); 364 for(int i = 1; i <= m; i++) { 365 scanf("%d%d%lld", &node[i].x, &node[i].y, &node[i].z); 366 if(node[i].x > node[i].y) std::swap(node[i].x, node[i].y); 367 else if(node[i].x == node[i].y) { 368 m--; 369 i--; 370 } 371 node[i].val = dis(node[i].x, node[i].y) - node[i].z; 372 //printf("node %d [%d %d] val = %lld ", i, node[i].x, node[i].y, node[i].val); 373 } 374 // solve 375 if(FLAG) { 376 Line::solve(); 377 clear(); 378 return; 379 } 380 if(c0) { 381 C0::solve(); 382 clear(); 383 return; 384 } 385 386 if(m > 300) { 387 clear(); 388 return; 389 } 390 391 LL ans = -INF; 392 for(int i = 1; i < m; i++) { 393 Add(node[i].x, node[i].y); 394 for(int j = i + 1; j <= m; j++) { 395 /// choose i j 396 LL Val = ask(node[j].x, node[j].y); 397 //printf("Val = %lld ", Val); 398 if(Val) { 399 ans = std::max(ans, node[i].val + node[j].val - Val); 400 } 401 } 402 tag[1] = sum[1] = 0; 403 } 404 if(ans == -INF) { 405 puts("F"); 406 } 407 else { 408 printf("%lld ", ans); 409 } 410 clear(); 411 return; 412 } 413 414 int main() { 415 416 int T; 417 scanf("%d", &T); 418 while(T--) solve(); 419 return 0; 420 }
接下来是这个东西...
然而我只得到了20分...可能比较丑。
只枚举有公共部分的链：考虑枚举每条边，并m2枚举经过这条边的链。
这样两条链可能会在多条边上算多次，于是我们限定某一个链的lca是这条边的上端点。
具体一点，线段树合并搞出经过每条边的链，然后枚举以这条边为lca的链，预处理树上距离。然后枚举经过这条边的所有链，并更新答案。
其实可以O(1)lca和长链剖分O(1)k级祖先来优化，但是我已经写不动啦......
1 /** 2 * There is no end though there is a start in space. ---Infinity. 3 * It has own power, it ruins, and it goes though there is a start also in the star. ---Finite. 4 * Only the person who was wisdom can read the most foolish one from the history. 5 * The fish that lives in the sea doesn't know the world in the land. 6 * It also ruins and goes if they have wisdom. 7 * It is funnier that man exceeds the speed of light than fish start living in the land. 8 * It can be said that this is an final ultimatum from the god to the people who can fight. 9 * 10 * Steins;Gate 11 */ 12 13 #include <bits/stdc++.h> 14 15 #define forson(x, i) for(int i = e[x]; i; i = edge[i].nex) 16 17 typedef long long LL; 18 const int N = 100010, M = 20000010; 19 const LL INF = 9e18; 20 21 struct Edge { 22 int nex, v; 23 LL len; 24 bool vis; 25 }edge[N << 1]; int tp = 1; 26 27 struct Node { 28 int x, y; 29 LL z, val; 30 inline bool operator < (const Node &w) const { 31 if(x != w.x) return x < w.x; 32 return y < w.y; 33 } 34 }node[N]; 35 36 int n, e[N], m, fa[N], deep[N], ls[M], rs[M], tot; 37 int num, pos[N], son[N], siz[N], top[N], rt[N], pw[N]; 38 LL Sum[N], sum[N << 2], d[N], tag[N << 2]; 39 40 inline void add(int x, int y, LL z) { 41 tp++; 42 edge[tp].v = y; 43 edge[tp].len = z; 44 edge[tp].vis = 0; 45 edge[tp].nex = e[x]; 46 e[x] = tp; 47 return; 48 } 49 50 void DFS_1(int x, int f) { /// get fa son siz d deep 51 fa[x] = f; 52 deep[x] = deep[f] + 1; 53 siz[x] = 1; 54 forson(x, i) { 55 int y = edge[i].v; 56 if(y == f) continue; 57 d[y] = d[x] + edge[i].len; 58 DFS_1(y, x); 59 siz[x] += siz[y]; 60 if(siz[y] > siz[son[x]]) { 61 son[x] = y; 62 } 63 } 64 return; 65 } 66 67 void DFS_2(int x, int f) { /// get top id pos 68 pos[x] = ++num; 69 top[x] = f; 70 Sum[num] = d[x] - d[fa[x]]; 71 if(son[x]) { 72 DFS_2(son[x], f); 73 } 74 forson(x, i) { 75 int y = edge[i].v; 76 if(y == fa[x] || y == son[x]) continue; 77 DFS_2(y, y); 78 } 79 return; 80 } 81 82 inline int lca(int x, int y) { 83 while(top[x] != top[y]) { 84 if(deep[top[x]] > deep[top[y]]) { 85 x = fa[top[x]]; 86 } 87 else { 88 y = fa[top[y]]; 89 } 90 } 91 return deep[x] < deep[y] ? x : y; 92 } 93 94 inline LL dis(int x, int y) { 95 return d[x] + d[y] - 2 * d[lca(x, y)]; 96 } 97 98 inline void pushdown(int l, int r, int o) { 99 if(tag[o] != -1) { 100 if(tag[o] == 0) { 101 sum[o << 1] = sum[o << 1 | 1] = 0; 102 } 103 else { 104 int mid = (l + r) >> 1; 105 sum[o << 1] = Sum[mid] - Sum[l - 1]; 106 sum[o << 1 | 1] = Sum[r] - Sum[mid]; 107 } 108 tag[o << 1] = tag[o << 1 | 1] = tag[o]; 109 tag[o] = -1; 110 } 111 return; 112 } 113 114 void Add(int L, int R, int l, int r, int o) { 115 if(L <= l && r <= R) { 116 sum[o] = Sum[r] - Sum[l - 1]; 117 tag[o] = 1; 118 return; 119 } 120 int mid = (l + r) >> 1; 121 pushdown(l, r, o); 122 if(L <= mid) Add(L, R, l, mid, o << 1); 123 if(mid < R) Add(L, R, mid + 1, r, o << 1 | 1); 124 sum[o] = sum[o << 1] + sum[o << 1 | 1]; 125 return; 126 } 127 128 LL ask(int L, int R, int l, int r, int o) { 129 if(L <= l && r <= R) { 130 return sum[o]; 131 } 132 int mid = (l + r) >> 1; 133 pushdown(l, r, o); 134 LL ans = 0; 135 if(L <= mid) { 136 ans += ask(L, R, l, mid, o << 1); 137 } 138 if(mid < R) { 139 ans += ask(L, R, mid + 1, r, o << 1 | 1); 140 } 141 return ans; 142 } 143 144 inline void Add(int x, int y) { 145 while(top[x] != top[y]) { 146 if(deep[top[x]] > deep[top[y]]) { 147 std::swap(x, y); 148 } 149 Add(pos[top[y]], pos[y], 1, n, 1); 150 y = fa[top[y]]; 151 } 152 if(deep[x] > deep[y]) std::swap(x, y); 153 if(x != y) Add(pos[x] + 1, pos[y], 1, n, 1); 154 return; 155 } 156 157 inline LL ask(int x, int y) { 158 LL ans = 0; 159 while(top[x] != top[y]) { 160 if(deep[top[x]] > deep[top[y]]) { 161 std::swap(x, y); 162 } 163 ans += ask(pos[top[y]], pos[y], 1, n, 1); 164 y = fa[top[y]]; 165 } 166 if(deep[x] > deep[y]) std::swap(x, y); 167 if(x != y) ans += ask(pos[x] + 1, pos[y], 1, n, 1); 168 return ans; 169 } 170 171 inline void clear() { 172 memset(e + 1, 0, n * sizeof(int)); 173 memset(son + 1, 0, n * sizeof(int)); 174 tp = 1; 175 num = 0; 176 return; 177 } 178 /* 179 1 180 5 181 1 2 1 182 2 3 3 183 3 4 3 184 1 5 9 185 2 186 1 5 5 187 2 3 8 188 */ 189 190 namespace Line { 191 LL ta[N], ta2[N]; 192 inline LL ask(int i) { 193 LL ans = -INF; 194 for(; i; i -= i & (-i)) { 195 ans = std::max(ans, ta[i]); 196 } 197 return ans; 198 } 199 inline void Add(int i, LL v) { 200 for(; i <= n; i += i & (-i)) { 201 ta[i] = std::max(ta[i], v); 202 } 203 return; 204 } 205 inline LL ask2(int i) { 206 LL ans = -INF; 207 for(; i; i -= i & (-i)) { 208 ans = std::max(ans, ta2[i]); 209 } 210 return ans; 211 } 212 inline void Add2(int i, LL v) { 213 for(; i <= n; i += i & (-i)) { 214 ta2[i] = std::max(ta2[i], v); 215 } 216 return; 217 } 218 inline void solve() { 219 for(int i = 1; i <= n; i++) ta2[i] = ta[i] = -INF; 220 LL ans = -INF; 221 std::sort(node + 1, node + m + 1); 222 for(int i = m; i >= 1; i--) { 223 LL temp = ask(node[i].y - 1); 224 if(temp != -INF) { 225 ans = std::max(ans, temp + node[i].val - Sum[node[i].y]); 226 } 227 temp = ask2(node[i].y); 228 if(temp != -INF) { 229 ans = std::max(ans, temp + node[i].val); 230 } 231 Add(node[i].x, node[i].val + Sum[node[i].x]); 232 Add2(node[i].y, node[i].val + Sum[node[i].x] - Sum[node[i].y]); 233 } 234 if(ans == -INF) puts("F"); 235 else { 236 printf("%lld ", ans); 237 } 238 return; 239 } 240 } 241 242 namespace C0 { 243 int xx; 244 LL Ans, X[N]; 245 std::vector<int> v[N]; 246 namespace seg { 247 int cnt[M]; 248 LL sum[M]; 249 void insert(int p, int v, int l, int r, int &o) { 250 if(!o) { 251 o = ++tot; 252 sum[o] = cnt[o] = ls[o] = rs[o] = 0; 253 } 254 if(l == r) { 255 sum[o] += v * X[r]; 256 cnt[o] += v; 257 return; 258 } 259 int mid = (l + r) >> 1; 260 if(p <= mid) { 261 insert(p, v, l, mid, ls[o]); 262 } 263 else { 264 insert(p, v, mid + 1, r, rs[o]); 265 } 266 sum[o] = sum[ls[o]] + sum[rs[o]]; 267 cnt[o] = cnt[ls[o]] + cnt[rs[o]]; 268 return; 269 } 270 LL ask(int k, int l, int r, int o) { 271 if(l == r) { 272 return k * X[r]; 273 } 274 int mid = (l + r) >> 1; 275 if(k <= cnt[rs[o]]) { 276 return ask(k, mid + 1, r, rs[o]); 277 } 278 else { 279 return ask(k - cnt[rs[o]], l, mid, ls[o]) + sum[rs[o]]; 280 } 281 } 282 inline LL Ask(int x) { 283 if(cnt[rt[x]] < 2) return -INF; 284 return ask(2, 1, xx, rt[x]); 285 } 286 int merge(int x, int y) { 287 if(!x || !y) return x | y; 288 sum[x] += sum[y]; 289 cnt[x] += cnt[y]; 290 ls[x] = merge(ls[x], ls[y]); 291 rs[x] = merge(rs[x], rs[y]); 292 return x; 293 } 294 inline void Merge(int x, int y) { 295 rt[x] = merge(rt[x], rt[y]); 296 return; 297 } 298 } 299 void DFS(int x, int f) { 300 forson(x, i) { 301 int y = edge[i].v; 302 if(y == f) continue; 303 DFS(y, x); 304 Ans = std::max(Ans, seg::Ask(y)); 305 seg::Merge(x, y); 306 } 307 for(int i = 0; i < (int)v[x].size(); i++) { 308 int t = v[x][i]; 309 if(t > 0) { 310 seg::insert(t, 1, 1, xx, rt[x]); 311 } 312 else { 313 seg::insert(-t, -1, 1, xx, rt[x]); 314 } 315 } 316 return; 317 } 318 inline void solve() { 319 memset(rt + 1, 0, n * sizeof(int)); 320 tot = 0; 321 Ans = -INF; 322 for(int i = 1; i <= n; i++) v[i].clear(); 323 for(int i = 1; i <= m; i++) { 324 X[i] = node[i].val; 325 } 326 std::sort(X + 1, X + m + 1); 327 xx = std::unique(X + 1, X + m + 1) - X - 1; 328 for(int i = 1; i <= m; i++) { 329 int val = std::lower_bound(X + 1, X + xx + 1, node[i].val) - X; 330 int x = node[i].x, y = node[i].y, z = lca(node[i].x, node[i].y); 331 v[x].push_back(val); 332 v[y].push_back(val); 333 v[z].push_back(-val); 334 v[z].push_back(-val); 335 } 336 DFS(1, 0); 337 if(Ans == -INF) puts("F"); 338 else printf("%lld ", Ans); 339 return; 340 } 341 } 342 343 namespace tbf { 344 LL d2[N], Ans; 345 int Top, stk[N], stk2[N], Top2, faa[N][20]; 346 std::vector<int> vx[N], v[N]; 347 LL dis2(int x, int y) { 348 return d2[x] + d2[y] - 2 * d2[lca(x, y)]; 349 } 350 void DFS_3(int x, int f) { /// get dis2 351 forson(x, i) { 352 int y = edge[i].v; 353 if(y == f) continue; 354 d2[y] = d2[x] + edge[i].len * edge[i].vis; 355 DFS_3(y, x); 356 } 357 return; 358 } 359 bool DFS_4(int x, int f, int t) { /// visit a link 360 if(x == t) { 361 Top = 0; 362 return true; 363 } 364 forson(x, i) { 365 int y = edge[i].v; 366 if(y == f) { 367 continue; 368 } 369 if(DFS_4(y, x, t)) { 370 edge[i].vis = edge[i ^ 1].vis = 1; 371 stk[++Top] = i; 372 return true; 373 } 374 } 375 return false; 376 } 377 namespace seg { 378 int sum[M]; 379 void insert(int p, int v, int l, int r, int &o) { 380 if(!o) { 381 o = ++tot; 382 ls[o] = rs[o] = sum[o] = 0; 383 } 384 if(l == r) { 385 sum[o] += v; 386 return; 387 } 388 int mid = (l + r) >> 1; 389 if(p <= mid) insert(p, v, l, mid, ls[o]); 390 else insert(p, v, mid + 1, r, rs[o]); 391 sum[o] = sum[ls[o]] + sum[rs[o]]; 392 return; 393 } 394 void solve(int l, int r, int o) { 395 if(l == r) { 396 stk2[++Top2] = r; 397 return; 398 } 399 int mid = (l + r) >> 1; 400 if(sum[ls[o]]) { 401 solve(l, mid, ls[o]); 402 } 403 if(sum[rs[o]]) { 404 solve(mid + 1, r, rs[o]); 405 } 406 return; 407 } 408 int merge(int x, int y) { 409 if(!x || !y) return x | y; 410 sum[x] += sum[y]; 411 ls[x] = merge(ls[x], ls[y]); 412 rs[x] = merge(rs[x], rs[y]); 413 return x; 414 } 415 } 416 417 void DFS_5(int x, int f) { 418 forson(x, i) { 419 int y = edge[i].v; 420 if(y == f) continue; 421 DFS_5(y, x); 422 Top2 = 0; 423 seg::solve(1, m, rt[y]); 424 for(int j = 0; j < (int)vx[y].size(); j++) { 425 int z = vx[y][j]; 426 /// link z 427 DFS_4(node[z].x, 0, node[z].y); 428 DFS_3(1, 0); 429 for(int k = 1; k <= Top2; k++) { 430 int w = stk2[k]; /// link w 431 if(w == z) continue; 432 Ans = std::max(Ans, node[z].val + node[w].val - dis2(node[w].x, node[w].y)); 433 } 434 while(Top) { 435 edge[stk[Top]].vis = edge[stk[Top] ^ 1].vis = 0; 436 Top--; 437 } 438 } 439 rt[x] = seg::merge(rt[x], rt[y]); 440 } 441 /// insert 442 for(int i = 0; i < (int)v[x].size(); i++) { 443 int t = v[x][i]; 444 if(t > 0) { 445 seg::insert(t, 1, 1, m, rt[x]); 446 } 447 else { 448 seg::insert(-t, -1, 1, m, rt[x]); 449 } 450 } 451 return; 452 } 453 void DFS_6(int x, int f) { 454 faa[x][0] = f; 455 forson(x, i) { 456 int y = edge[i].v; 457 if(y == f) continue; 458 DFS_6(y, x); 459 } 460 return; 461 } 462 inline void prework() { 463 for(int j = 1; j <= pw[n]; j++) { 464 for(int i = 1; i <= n; i++) { 465 faa[i][j] = faa[faa[i][j - 1]][j - 1]; 466 } 467 } 468 return; 469 } 470 inline int Find(int x, int z) { 471 int t = pw[n]; 472 while(t >= 0 && fa[x] != z) { 473 if(deep[faa[x][t]] > deep[z]) { 474 x = faa[x][t]; 475 } 476 t--; 477 } 478 return x; 479 } 480 inline void solve() { 481 tot = 0; 482 Ans = -INF; 483 memset(rt + 1, 0, n * sizeof(int)); 484 for(int i = 1; i <= n; i++) { 485 v[i].clear(); 486 vx[i].clear(); 487 } 488 DFS_6(1, 0); 489 prework(); 490 for(int i = 1; i <= m; i++) { 491 int x = node[i].x, y = node[i].y, z = lca(x, y); 492 v[x].push_back(i); 493 v[y].push_back(i); 494 v[z].push_back(-i); 495 v[z].push_back(-i); 496 if(x != z) { 497 int xx = Find(x, z); 498 vx[xx].push_back(i); 499 } 500 if(y != z) { 501 int yy = Find(y, z); 502 vx[yy].push_back(i); 503 } 504 } 505 DFS_5(1, 0); 506 if(Ans == -INF) puts("F"); 507 else printf("%lld ", Ans); 508 return; 509 } 510 } /** namespace tbf */ 511 512 inline void solve() { 513 514 sum[1] = tag[1] = 0; 515 516 scanf("%d", &n); 517 LL z; 518 bool FLAG = 1, c0 = 1; 519 for(int i = 1, x, y; i < n; i++) { 520 scanf("%d%d%lld", &x, &y, &z); 521 add(x, y, z); 522 add(y, x, z); 523 if(z) c0 = 0; 524 if(y != x + 1) { 525 FLAG = 0; 526 } 527 } 528 529 DFS_1(1, 0); 530 DFS_2(1, 1); 531 532 for(int i = 1; i <= n; i++) { 533 Sum[i] += Sum[i - 1]; 534 } 535 536 scanf("%d", &m); 537 for(int i = 1; i <= m; i++) { 538 scanf("%d%d%lld", &node[i].x, &node[i].y, &node[i].z); 539 if(node[i].x > node[i].y) std::swap(node[i].x, node[i].y); 540 else if(node[i].x == node[i].y) { 541 m--; 542 i--; 543 } 544 node[i].val = dis(node[i].x, node[i].y) - node[i].z; 545 //printf("node %d [%d %d] val = %lld ", i, node[i].x, node[i].y, node[i].val); 546 } 547 // solve 548 549 tbf::solve(); 550 clear(); 551 return; 552 553 if(FLAG) { 554 Line::solve(); 555 clear(); 556 return; 557 } 558 if(c0) { 559 C0::solve(); 560 clear(); 561 return; 562 } 563 564 if(m > 300) { 565 tbf::solve(); 566 clear(); 567 return; 568 } 569 570 LL ans = -INF; 571 for(int i = 1; i < m; i++) { 572 Add(node[i].x, node[i].y); 573 for(int j = i + 1; j <= m; j++) { 574 /// choose i j 575 LL Val = ask(node[j].x, node[j].y); 576 //printf("Val = %lld ", Val); 577 if(Val) { 578 ans = std::max(ans, node[i].val + node[j].val - Val); 579 } 580 } 581 tag[1] = sum[1] = 0; 582 } 583 if(ans == -INF) { 584 puts("F"); 585 } 586 else { 587 printf("%lld ", ans); 588 } 589 clear(); 590 return; 591 } 592 593 int main() { 594 595 for(int i = 2; i < N; i++) pw[i] = pw[i >> 1] + 1; 596 597 int T; 598 scanf("%d", &T); 599 while(T--) solve(); 600 return 0; 601 }
接下来是S1的部分分：所有lca全部不同。
实际上(参考yww的代码)我们要维护两个东西，一个是val最大值，一个是val + d[lca]的最大值。然后更新的时候，把线段树交错更新。
也就是说左儿子(较浅)的val + 右儿子(较深)的(val + d[lca])然后减去红点的深度。
然后我们要在插入和merge的同时更新。注意到我们还有一个删除操作，x子树搞完之后，深度大于等于x的全部失效，也就是我们要把线段树右边的一段全部清除。
DFS那里插入删除merge的顺序非常重要....反正我试了很久才搞对顺序。
1 #include <bits/stdc++.h> 2 3 #define forson(x, i) for(int i = e[x]; i; i = edge[i].nex) 4 5 typedef long long LL; 6 const int N = 100010; 7 const LL INF = 4e18; 8 9 struct Edge { 10 int nex, v; 11 LL len; 12 }edge[N << 1]; int tp = 1; 13 14 struct Node { 15 int x, y, w; 16 LL val, z; 17 }node[N]; 18 19 int e[N], num2, pos2[N], ST[N][20], pw[N], deep[N], n, m, fa[N]; 20 LL d[N]; 21 std::bitset<N> visLca; 22 23 inline void add(int x, int y, LL z) { 24 tp++; 25 edge[tp].v = y; 26 edge[tp].len = z; 27 edge[tp].nex = e[x]; 28 e[x] = tp; 29 return; 30 } 31 32 inline void clear() { 33 memset(e + 1, 0, n * sizeof(int)); 34 visLca.reset(); 35 tp = 1; 36 num2 = 0; 37 return; 38 } 39 40 void DFS(int x, int f) { 41 pos2[x] = ++num2; 42 ST[num2][0] = x; 43 deep[x] = deep[f] + 1; 44 fa[x] = f; 45 forson(x, i) { 46 int y = edge[i].v; 47 if(y == f) { 48 continue; 49 } 50 d[y] = d[x] + edge[i].len; 51 DFS(y, x); 52 ST[++num2][0] = x; 53 } 54 return; 55 } 56 57 inline void prework() { 58 for(int j = 1; j <= pw[num2]; j++) { 59 for(int i = 1; i + (1 << j) - 1 <= num2; i++) { 60 if(deep[ST[i][j - 1]] < deep[ST[i + (1 << (j - 1))][j - 1]]) { 61 ST[i][j] = ST[i][j - 1]; 62 } 63 else { 64 ST[i][j] = ST[i + (1 << (j - 1))][j - 1]; 65 } 66 } 67 } 68 return; 69 } 70 71 inline int lca(int x, int y) { 72 x = pos2[x]; 73 y = pos2[y]; 74 if(x > y) std::swap(x, y); 75 int t = pw[y - x + 1]; 76 if(deep[ST[x][t]] < deep[ST[y - (1 << t) + 1][t]]) { 77 return ST[x][t]; 78 } 79 else { 80 return ST[y - (1 << t) + 1][t]; 81 } 82 } 83 84 inline LL dis(int x, int y) { 85 return d[x] + d[y] - 2 * d[lca(x, y)]; 86 } 87 88 namespace Sone { /// lca != lca 89 LL Ans; 90 int rt[N]; 91 std::vector<int> v[N]; 92 namespace seg { 93 const int M = 10000010; 94 95 int ls[M], rs[M], tot; 96 LL large[M], largeD[M]; 97 98 inline void pushup(int o) { 99 large[o] = std::max(large[ls[o]], large[rs[o]]); 100 largeD[o] = std::max(largeD[ls[o]], largeD[rs[o]]); 101 return; 102 } 103 void insert(int p, LL v, LL v2, LL v3, int l, int r, int &o) { 104 if(!o) { 105 o = ++tot; 106 ls[o] = rs[o] = 0; 107 large[o] = largeD[o] = -INF; 108 } 109 if(l == r) { 110 large[o] = std::max(large[o], v); 111 largeD[o] = std::max(largeD[o], v + v2); 112 return; 113 } 114 int mid = (l + r) >> 1; 115 if(p <= mid) { 116 Ans = std::max(Ans, largeD[rs[o]] + v - v3); 117 insert(p, v, v2, v3, l, mid, ls[o]); 118 } 119 else { 120 Ans = std::max(Ans, large[ls[o]] + v + v2 - v3); 121 insert(p, v, v2, v3, mid + 1, r, rs[o]); 122 } 123 pushup(o); 124 return; 125 } 126 int merge(int x, int y, LL v) { 127 //printf("merge x = %d y = %d v = %lld ", x, y, v); 128 if(!x || !y) return x | y; 129 Ans = std::max(Ans, std::max(large[ls[x]] + largeD[rs[y]], large[ls[y]] + largeD[rs[x]]) - v); 130 large[x] = std::max(large[x], large[y]); 131 largeD[x] = std::max(largeD[x], largeD[y]); 132 ls[x] = merge(ls[x], ls[y], v); 133 rs[x] = merge(rs[x], rs[y], v); 134 return x; 135 } 136 void del(int p, int l, int r, int o) { 137 if(!o) { 138 return; 139 } 140 if(l == r) { 141 large[o] = largeD[o] = -INF; 142 return; 143 } 144 int mid = (l + r) >> 1; 145 if(p <= mid) { 146 large[rs[o]] = largeD[rs[o]] = -INF; 147 del(p, l, mid, ls[o]); 148 } 149 else { 150 del(p, mid + 1, r, rs[o]); 151 } 152 pushup(o); 153 return; 154 } 155 } 156 157 void DFS_1(int x, int f) { 158 //printf("x = %d ", x); 159 for(int i = 0; i < (int)v[x].size(); i++) { 160 int t = v[x][i]; 161 //printf("x = %d insert p = %d w = %d v1 = %lld v2 = %lld ", x, deep[node[t].w], node[t].w, node[t].val, d[node[t].w]); 162 seg::insert(deep[node[t].w], node[t].val, d[node[t].w], d[x], 1, n, rt[x]); 163 } 164 forson(x, i) { 165 int y = edge[i].v; 166 if(y == f) continue; 167 //printf("y = %d ", y); 168 DFS_1(y, x); 169 //printf("merge %d -> %d ", x, y); 170 seg::del(deep[x], 1, n, rt[y]); 171 rt[x] = seg::merge(rt[x], rt[y], d[x]); 172 //printf("Ans = %lld ", Ans); 173 } 174 return; 175 } 176 177 inline void solve() { 178 seg::tot = 0; 179 seg::large[0] = seg::largeD[0] = -INF; 180 Ans = -INF; 181 memset(rt + 1, 0, n * sizeof(int)); 182 for(int i = 1; i <= n; i++) v[i].clear(); 183 184 for(int i = 1; i <= m; i++) { 185 int x = node[i].x, y = node[i].y, z = node[i].w; 186 if(x != z) { 187 v[x].push_back(i); 188 } 189 if(y != z) { 190 v[y].push_back(i); 191 } 192 } 193 194 DFS_1(1, 0); 195 if(Ans < (LL)(-3e18)) puts("F"); 196 else printf("%lld ", Ans); 197 return; 198 } 199 } 200 201 namespace Stwo { /// lca = 1 202 inline void solve() { 203 204 return; 205 } 206 } 207 208 inline void solve() { 209 scanf("%d", &n); 210 LL z; 211 for(int i = 1, x, y; i < n; i++) { 212 scanf("%d%d%lld", &x, &y, &z); 213 add(x, y, z); 214 add(y, x, z); 215 } 216 217 DFS(1, 0); 218 prework(); 219 220 scanf("%d", &m); 221 bool S1 = true, S2 = true; 222 int lastLca = -1; 223 for(int i = 1, x, y; i <= m; i++) { 224 scanf("%d%d%lld", &x, &y, &z); 225 node[i].x = x; 226 node[i].y = y; 227 node[i].z = z; 228 node[i].val = dis(x, y) - z; 229 int temp = lca(x, y); 230 if(lastLca == -1) lastLca = temp; 231 else if(lastLca != temp) S2 = false; 232 if(visLca[temp]) S1 = false; 233 else visLca.set(temp); 234 node[i].w = temp; 235 } 236 237 if(S1) { 238 Sone::solve(); 239 return; 240 } 241 if(S2) { 242 Stwo::solve(); 243 return; 244 } 245 246 return; 247 } 248 249 int main() { 250 251 freopen("in.in", "r", stdin); 252 freopen("my.out", "w", stdout); 253 254 for(int i = 2; i < N; i++) pw[i] = pw[i >> 1] + 1; 255 256 int T; 257 scanf("%d", &T); 258 while(T--) { 259 solve(); 260 if(T) clear(); 261 } 262 return 0; 263 }
接下来是S2的lca全部为1的部分分。
继续抄题解。
我的理解是，首先把2 * 总价值转化成链长之和 - 代价之和 + 蓝点距离 + 绿点距离。我们尝试新建两个点来表示出这个东西。
首先把蓝点距离转化成蓝a深度 + 蓝b深度 - 2 * 红点深度。
对于一条链a，我们在绿a处挂上一个节点绿a'，使得它到绿a的距离为链长 - 代价 + 蓝a深度。
然后我们发现，这样两个附加点的距离正好凑出了上面那个2 * 总价值，只要再减去2 * 红点深度就行了。
有这个结论是因为边权非负。那么我们新加的边不是有一堆负权吗？为什么成立呢？
因为我们可以给这些新加的边(假装加上了)加上一个很大的正值来达到非负。为什么这个方法在一般的树上不适用呢？因为我们不知道直径上有多少条边加了这个大正值。但是本题中会有恰两条边加上这个正值。
所以做法就出来了。具体来说，我们DFS这棵树，然后在每个蓝a处把绿a'加入点集，同理在绿a处把蓝a'加入点集。
然后合并点集的同时更新答案。注意当前点为根的时候不能更新，因为可能会把蓝a'和绿a'的距离算入答案。也没必要更新，因为根不可能作为红点。
注意只需要挂可能成为蓝a的对应的绿a'，就是说一条链直上直下的时候有一个点不用加。
1 #include <bits/stdc++.h> 2 3 #define forson(x, i) for(int i = e[x]; i; i = edge[i].nex) 4 5 using namespace std; 6 7 typedef long long LL; 8 const int N = 300010; 9 const LL INF = 4e18; 10 11 struct Edge { 12 int nex, v; 13 LL len; 14 }edge[N << 1]; int tp = 1; 15 16 struct Node { 17 int x, y, w; 18 LL val, z; 19 }node[N]; 20 21 struct Data { 22 int a, b; 23 Data(int A = 0, int B = 0) : a(A), b(B) {} 24 }data[N]; 25 26 int e[N], num2, pos2[N], ST[N][20], pw[N], deep[N], n, m, fa[N], fr[N]; 27 LL d[N]; 28 std::bitset<N> visLca; 29 30 inline void add(int x, int y, LL z) { 31 tp++; 32 edge[tp].v = y; 33 edge[tp].len = z; 34 edge[tp].nex = e[x]; 35 e[x] = tp; 36 return; 37 } 38 39 inline void clear() { 40 memset(e + 1, 0, n * sizeof(int)); 41 visLca.reset(); 42 tp = 1; 43 num2 = 0; 44 return; 45 } 46 47 void DFS(int x, int f) { 48 pos2[x] = ++num2; 49 ST[num2][0] = x; 50 deep[x] = deep[f] + 1; 51 fa[x] = f; 52 forson(x, i) { 53 int y = edge[i].v; 54 if(y == f) { 55 continue; 56 } 57 d[y] = d[x] + edge[i].len; 58 DFS(y, x); 59 ST[++num2][0] = x; 60 } 61 return; 62 } 63 64 inline void prework() { 65 for(int j = 1; j <= pw[num2]; j++) { 66 for(int i = 1; i + (1 << j) - 1 <= num2; i++) { 67 if(deep[ST[i][j - 1]] < deep[ST[i + (1 << (j - 1))][j - 1]]) { 68 ST[i][j] = ST[i][j - 1]; 69 } 70 else { 71 ST[i][j] = ST[i + (1 << (j - 1))][j - 1]; 72 } 73 } 74 } 75 return; 76 } 77 78 inline int lca(int x, int y) { 79 x = pos2[fr[x]]; 80 y = pos2[fr[y]]; 81 if(x > y) std::swap(x, y); 82 int t = pw[y - x + 1]; 83 if(deep[ST[x][t]] < deep[ST[y - (1 << t) + 1][t]]) { 84 return ST[x][t]; 85 } 86 else { 87 return ST[y - (1 << t) + 1][t]; 88 } 89 } 90 91 inline LL dis(int x, int y) { 92 if(!x || !y) return -INF; 93 if(x > n && (x - n + 1) / 2 == (y - n + 1) / 2) return -INF; 94 //if(x > n) printf("dis %d %d = %lld ", x, y, d[x] + d[y] - 2 * d[lca(x, y)]); 95 return d[x] + d[y] - 2 * d[lca(x, y)]; 96 } 97 98 namespace Sone { /// lca != lca 99 LL Ans; 100 int rt[N]; 101 std::vector<int> v[N]; 102 namespace seg { 103 const int M = 10000010; 104 105 int ls[M], rs[M], tot; 106 LL large[M], largeD[M]; 107 108 inline void pushup(int o) { 109 large[o] = std::max(large[ls[o]], large[rs[o]]); 110 largeD[o] = std::max(largeD[ls[o]], largeD[rs[o]]); 111 return; 112 } 113 void insert(int p, LL v, LL v2, LL v3, int l, int r, int &o) { 114 if(!o) { 115 o = ++tot; 116 ls[o] = rs[o] = 0; 117 large[o] = largeD[o] = -INF; 118 } 119 if(l == r) { 120 large[o] = std::max(large[o], v); 121 largeD[o] = std::max(largeD[o], v + v2); 122 return; 123 } 124 int mid = (l + r) >> 1; 125 if(p <= mid) { 126 Ans = std::max(Ans, largeD[rs[o]] + v - v3); 127 insert(p, v, v2, v3, l, mid, ls[o]); 128 } 129 else { 130 Ans = std::max(Ans, large[ls[o]] + v + v2 - v3); 131 insert(p, v, v2, v3, mid + 1, r, rs[o]); 132 } 133 pushup(o); 134 return; 135 } 136 int merge(int x, int y, LL v) { 137 //printf("merge x = %d y = %d v = %lld ", x, y, v); 138 if(!x || !y) return x | y; 139 Ans = std::max(Ans, std::max(large[ls[x]] + largeD[rs[y]], large[ls[y]] + largeD[rs[x]]) - v); 140 large[x] = std::max(large[x], large[y]); 141 largeD[x] = std::max(largeD[x], largeD[y]); 142 ls[x] = merge(ls[x], ls[y], v); 143 rs[x] = merge(rs[x], rs[y], v); 144 return x; 145 } 146 void del(int p, int l, int r, int o) { 147 if(!o) { 148 return; 149 } 150 if(l == r) { 151 large[o] = largeD[o] = -INF; 152 return; 153 } 154 int mid = (l + r) >> 1; 155 if(p <= mid) { 156 large[rs[o]] = largeD[rs[o]] = -INF; 157 del(p, l, mid, ls[o]); 158 } 159 else { 160 del(p, mid + 1, r, rs[o]); 161 } 162 pushup(o); 163 return; 164 } 165 } 166 167 void DFS_1(int x, int f) { 168 //printf("x = %d ", x); 169 for(int i = 0; i < (int)v[x].size(); i++) { 170 int t = v[x][i]; 171 //printf("x = %d insert p = %d w = %d v1 = %lld v2 = %lld ", x, deep[node[t].w], node[t].w, node[t].val, d[node[t].w]); 172 seg::insert(deep[node[t].w], node[t].val, d[node[t].w], d[x], 1, n, rt[x]); 173 } 174 forson(x, i) { 175 int y = edge[i].v; 176 if(y == f) continue; 177 //printf("y = %d ", y); 178 DFS_1(y, x); 179 //printf("merge %d -> %d ", x, y); 180 seg::del(deep[x], 1, n, rt[y]); 181 rt[x] = seg::merge(rt[x], rt[y], d[x]); 182 //printf("Ans = %lld ", Ans); 183 } 184 return; 185 } 186 187 inline void solve() { 188 seg::tot = 0; 189 seg::large[0] = seg::largeD[0] = -INF; 190 Ans = -INF; 191 memset(rt + 1, 0, n * sizeof(int)); 192 for(int i = 1; i <= n; i++) v[i].clear(); 193 194 for(int i = 1; i <= m; i++) { 195 int x = node[i].x, y = node[i].y, z = node[i].w; 196 if(x != z) { 197 v[x].push_back(i); 198 } 199 if(y != z) { 200 v[y].push_back(i); 201 } 202 } 203 204 DFS_1(1, 0); 205 if(Ans < (LL)(-3e18)) puts("F"); 206 else printf("%lld ", Ans); 207 return; 208 } 209 } 210 211 namespace Stwo { /// lca = 1 212 /*Edge EDGE[N << 1]; int TP; 213 int E[N], use[N], Time, fr[N]; 214 inline void ADD(int x, int y, LL z) { 215 TP++; 216 EDGE[TP].v = y; 217 EDGE[TP].len = z; 218 EDGE[TP].nex = E[x]; 219 E[x] = TP; 220 return; 221 } 222 inline void work(int x) { 223 if(use[x] != Time) { 224 use[x] = Time; 225 E[x] = 0; 226 ? = ?; 227 } 228 return; 229 }*/ 230 std::vector<int> v[N]; 231 LL Ans; 232 233 inline void update(Data &x, const Data &y, const LL &v) { 234 //printf("update v = %lld ", v); 235 LL d12 = dis(x.a, x.b), d34 = dis(y.a, y.b); 236 LL d13 = dis(x.a, y.a), d14 = dis(x.a, y.b); 237 LL d23 = dis(x.b, y.a), d24 = dis(x.b, y.b); 238 Ans = std::max(Ans, std::max(std::max(d13, d14), std::max(d23, d24)) - v); 239 int xa = x.a, xb = x.b; 240 if(d34 > d12) { 241 x = y; 242 d12 = d34; 243 } 244 if(d13 > d12) { 245 x = Data(xa, y.a); 246 d12 = d13; 247 } 248 if(d14 > d12) { 249 x = Data(xa, y.b); 250 d12 = d14; 251 } 252 if(d23 > d12) { 253 x = Data(xb, y.a); 254 d12 = d23; 255 } 256 if(d24 > d12) { 257 x = Data(xb, y.b); 258 } 259 if(!x.a) x = y; 260 //printf("Ans = %lld ", Ans); 261 return; 262 } 263 264 inline void update(Data &x, const int &a, const LL &v) { 265 //printf("insert v = %lld ", v); 266 LL d1 = dis(x.a, a), d2 = dis(x.b, a), d3 = dis(x.a, x.b); 267 Ans = std::max(Ans, std::max(d1, d2) - v); 268 int xa = x.a, xb = x.b; 269 if(d1 > d3) { 270 x.b = a; 271 d3 = d1; 272 } 273 if(d2 > d3) { 274 x = Data(xb, a); 275 } 276 if(!x.a) x.a = a; 277 //printf("Ans = %lld ", Ans); 278 return; 279 } 280 281 void DFS_1(int x, int f) { 282 forson(x, i) { 283 int y = edge[i].v; 284 if(y == f) continue; 285 DFS_1(y, x); 286 if(f) update(data[x], data[y], d[x] << 1); 287 //printf("data x : %d %d ", data[x].a, data[x].b); 288 } 289 if(f) { 290 for(int i = 0; i < (int)v[x].size(); i++) { 291 int t = v[x][i]; 292 update(data[x], t, d[x] << 1); 293 } 294 } 295 return; 296 } 297 298 inline void solve() { 299 300 //printf("S2 inside "); 301 302 for(int i = 1; i <= n; i++) { 303 v[i].clear(); 304 v[i].resize(0); 305 data[i].a = data[i].b = 0; 306 } 307 Ans = -INF; 308 for(int i = 1; i <= m; i++) { 309 int x = node[i].x, y = node[i].y, z = node[i].w; 310 if(x != z) { 311 fr[n + i * 2 - 1] = y; 312 d[n + i * 2 - 1] = d[y] + d[x] + node[i].val - node[i].z; 313 v[x].push_back(n + i * 2 - 1); 314 //printf("%d : d = %lld fr = %d ", n + i * 2 - 1, d[n + i * 2 - 1], fr[n + i * 2 - 1]); 315 } 316 if(y != z) { 317 fr[n + i * 2] = x; 318 d[n + i * 2] = d[x] + d[y] + node[i].val - node[i].z; 319 v[y].push_back(n + i * 2); 320 //printf("%d : d = %lld fr = %d ", n + i * 2, d[n + i * 2], fr[n + i * 2]); 321 } 322 } 323 //puts(""); 324 DFS_1(1, 0); 325 326 if(Ans < (LL)-3e18) puts("F"); 327 else printf("%lld ", Ans >> 1); 328 return; 329 } 330 } 331 332 inline void solve() { 333 scanf("%d", &n); 334 for(int i = 1; i <= n; i++) { 335 fr[i] = i; 336 } 337 LL z; 338 for(int i = 1, x, y; i < n; i++) { 339 scanf("%d%d%lld", &x, &y, &z); 340 add(x, y, z); 341 add(y, x, z); 342 } 343 344 DFS(1, 0); 345 prework(); 346 347 scanf("%d", &m); 348 bool S1 = true, S2 = true; 349 int lastLca = -1; 350 for(int i = 1, x, y; i <= m; i++) { 351 scanf("%d%d%lld", &x, &y, &z); 352 node[i].x = x; 353 node[i].y = y; 354 node[i].z = z; 355 node[i].val = dis(x, y) - z; 356 int temp = lca(x, y); 357 if(lastLca == -1) lastLca = temp; 358 else if(lastLca != temp) S2 = false; 359 if(visLca[temp]) S1 = false; 360 else visLca.set(temp); 361 node[i].w = temp; 362 } 363 364 if(S1) { 365 Sone::solve(); 366 return; 367 } 368 if(S2) { 369 Stwo::solve(); 370 return; 371 } 372 373 return; 374 } 375 376 int main() { 377 378 //freopen("in.in", "r", stdin); 379 //freopen("my.out", "w", stdout); 380 381 for(int i = 2; i < N; i++) pw[i] = pw[i >> 1] + 1; 382 383 int T; 384 scanf("%d", &T); 385 while(T--) { 386 solve(); 387 if(T) clear(); 388 } 389 return 0; 390 }
终于到正解了...考虑两条链如果有交，那么要么是直上直下的交，lca不同；要么是交了一段可能不直上直下的，lca相同。
那么lca不同的所有情况直接调用S1即可。lca相同的情况，可能在很多点相同。此时我们枚举每个点，把以这些点为lca的链都提出来建虚树，套用S2即可。
考虑为什么套用S1的时候不会把本来属于S2的情况算得更优。
注意到一个细节，就是S1中如果两个链加到线段树的同一下标上(lca相同)，我们是不会把这两条链拿去更新答案的。于是S1中更新的就都是lca不同的啦！
1 #include <bits/stdc++.h> 2 3 #define forson(x, i) for(int i = e[x]; i; i = edge[i].nex) 4 5 typedef long long LL; 6 const int N = 300010; 7 const LL INF = 4e18; 8 9 struct Edge { 10 int nex, v; 11 LL len; 12 }edge[N << 1]; int tp = 1; 13 14 struct Node { 15 int x, y, w; 16 LL val, z; 17 }node[N]; 18 19 struct Data { 20 int a, b; 21 Data(int A = 0, int B = 0) : a(A), b(B) {} 22 }data[N]; 23 24 int e[N], num2, pos2[N], ST[N][20], pw[N], deep[N], n, m, fa[N], fr[N]; 25 LL d[N], Ans; 26 27 inline void add(int x, int y, LL z) { 28 tp++; 29 edge[tp].v = y; 30 edge[tp].len = z; 31 edge[tp].nex = e[x]; 32 e[x] = tp; 33 return; 34 } 35 36 inline void clear() { 37 memset(e + 1, 0, n * sizeof(int)); 38 tp = 1; 39 num2 = 0; 40 return; 41 } 42 43 void DFS(int x, int f) { 44 pos2[x] = ++num2; 45 ST[num2][0] = x; 46 deep[x] = deep[f] + 1; 47 fa[x] = f; 48 forson(x, i) { 49 int y = edge[i].v; 50 if(y == f) { 51 continue; 52 } 53 d[y] = d[x] + edge[i].len; 54 DFS(y, x); 55 ST[++num2][0] = x; 56 } 57 return; 58 } 59 60 inline void prework() { 61 for(int j = 1; j <= pw[num2]; j++) { 62 for(int i = 1; i + (1 << j) - 1 <= num2; i++) { 63 if(deep[ST[i][j - 1]] < deep[ST[i + (1 << (j - 1))][j - 1]]) { 64 ST[i][j] = ST[i][j - 1]; 65 } 66 else { 67 ST[i][j] = ST[i + (1 << (j - 1))][j - 1]; 68 } 69 } 70 } 71 return; 72 } 73 74 inline int lca(int x, int y) { 75 x = pos2[fr[x]]; 76 y = pos2[fr[y]]; 77 if(x > y) std::swap(x, y); 78 int t = pw[y - x + 1]; 79 if(deep[ST[x][t]] < deep[ST[y - (1 << t) + 1][t]]) { 80 return ST[x][t]; 81 } 82 else { 83 return ST[y - (1 << t) + 1][t]; 84 } 85 } 86 87 inline LL dis(int x, int y) { 88 if(!x || !y) return -INF; 89 return d[x] + d[y] - 2 * d[lca(x, y)]; 90 } 91 92 inline bool cmp(const int &a, const int &b) { 93 return pos2[a] < pos2[b]; 94 } 95 96 namespace Sone { /// lca != lca 97 int rt[N]; 98 std::vector<int> v[N]; 99 namespace seg { 100 const int M = 10000010; 101 102 int ls[M], rs[M], tot; 103 LL large[M], largeD[M]; 104 105 inline void pushup(int o) { 106 large[o] = std::max(large[ls[o]], large[rs[o]]); 107 largeD[o] = std::max(largeD[ls[o]], largeD[rs[o]]); 108 return; 109 } 110 void insert(int p, LL v, LL v2, LL v3, int l, int r, int &o) { 111 if(!o) { 112 o = ++tot; 113 ls[o] = rs[o] = 0; 114 large[o] = largeD[o] = -INF; 115 } 116 if(l == r) { 117 large[o] = std::max(large[o], v); 118 largeD[o] = std::max(largeD[o], v + v2); 119 return; 120 } 121 int mid = (l + r) >> 1; 122 if(p <= mid) { 123 Ans = std::max(Ans, largeD[rs[o]] + v - v3); 124 insert(p, v, v2, v3, l, mid, ls[o]); 125 } 126 else { 127 Ans = std::max(Ans, large[ls[o]] + v + v2 - v3); 128 insert(p, v, v2, v3, mid + 1, r, rs[o]); 129 } 130 pushup(o); 131 return; 132 } 133 int merge(int x, int y, LL v) { 134 if(!x || !y) return x | y; 135 Ans = std::max(Ans, std::max(large[ls[x]] + largeD[rs[y]], large[ls[y]] + largeD[rs[x]]) - v); 136 large[x] = std::max(large[x], large[y]); 137 largeD[x] = std::max(largeD[x], largeD[y]); 138 ls[x] = merge(ls[x], ls[y], v); 139 rs[x] = merge(rs[x], rs[y], v); 140 return x; 141 } 142 void del(int p, int l, int r, int o) { 143 if(!o) { 144 return; 145 } 146 if(l == r) { 147 large[o] = largeD[o] = -INF; 148 return; 149 } 150 int mid = (l + r) >> 1; 151 if(p <= mid) { 152 large[rs[o]] = largeD[rs[o]] = -INF; 153 del(p, l, mid, ls[o]); 154 } 155 else { 156 del(p, mid + 1, r, rs[o]); 157 } 158 pushup(o); 159 return; 160 } 161 } 162 163 void DFS_1(int x, int f) { 164 for(int i = 0; i < (int)v[x].size(); i++) { 165 int t = v[x][i]; 166 seg::insert(deep[node[t].w], node[t].val, d[node[t].w], d[x], 1, n, rt[x]); 167 } 168 forson(x, i) { 169 int y = edge[i].v; 170 if(y == f) continue; 171 DFS_1(y, x); 172 seg::del(deep[x], 1, n, rt[y]); 173 rt[x] = seg::merge(rt[x], rt[y], d[x]); 174 } 175 return; 176 } 177 178 inline void solve() { 179 seg::tot = 0; 180 seg::large[0] = seg::largeD[0] = -INF; 181 memset(rt + 1, 0, n * sizeof(int)); 182 for(int i = 1; i <= n; i++) v[i].clear(); 183 184 for(int i = 1; i <= m; i++) { 185 int x = node[i].x, y = node[i].y, z = node[i].w; 186 if(x != z) { 187 v[x].push_back(i); 188 } 189 if(y != z) { 190 v[y].push_back(i); 191 } 192 } 193 194 DFS_1(1, 0); 195 return; 196 } 197 } 198 199 namespace Stwo { /// lca = 1 200 Edge EDGE[N << 1]; int TP; 201 int E[N], use[N], Time, cnt, imp[N], stk[N], top; 202 std::vector<int> v[N], vx[N]; 203 204 inline void ADD(int x, int y) { 205 TP++; 206 EDGE[TP].v = y; 207 EDGE[TP].nex = E[x]; 208 E[x] = TP; 209 return; 210 } 211 212 inline void work(int x) { 213 if(use[x] != Time) { 214 use[x] = Time; 215 E[x] = 0; 216 v[x].clear(); 217 data[x].a = data[x].b = 0; 218 } 219 return; 220 } 221 222 inline void update(Data &x, const Data &y, const LL &v) { 223 LL d12 = dis(x.a, x.b), d34 = dis(y.a, y.b); 224 LL d13 = dis(x.a, y.a), d14 = dis(x.a, y.b); 225 LL d23 = dis(x.b, y.a), d24 = dis(x.b, y.b); 226 Ans = std::max(Ans, (std::max(std::max(d13, d14), std::max(d23, d24)) - v) >> 1); 227 int xa = x.a, xb = x.b; 228 if(d34 > d12) { 229 x = y; 230 d12 = d34; 231 } 232 if(d13 > d12) { 233 x = Data(xa, y.a); 234 d12 = d13; 235 } 236 if(d14 > d12) { 237 x = Data(xa, y.b); 238 d12 = d14; 239 } 240 if(d23 > d12) { 241 x = Data(xb, y.a); 242 d12 = d23; 243 } 244 if(d24 > d12) { 245 x = Data(xb, y.b); 246 } 247 if(!x.a) x = y; 248 return; 249 } 250 251 inline void update(Data &x, const int &a, const LL &v) { 252 LL d1 = dis(x.a, a), d2 = dis(x.b, a), d3 = dis(x.a, x.b); 253 Ans = std::max(Ans, (std::max(d1, d2) - v) >> 1); 254 int xa = x.a, xb = x.b; 255 if(d1 > d3) { 256 x.b = a; 257 d3 = d1; 258 } 259 if(d2 > d3) { 260 x = Data(xb, a); 261 } 262 if(!x.a) x.a = a; 263 return; 264 } 265 266 void dfs(int x, int f) { 267 for(int i = E[x]; i; i = EDGE[i].nex) { 268 int y = EDGE[i].v; 269 if(y == f) continue; 270 dfs(y, x); 271 if(f) update(data[x], data[y], d[x] << 1); 272 } 273 if(f) { 274 for(int i = 0; i < (int)v[x].size(); i++) { 275 int t = v[x][i]; 276 update(data[x], t, d[x] << 1); 277 } 278 } 279 return; 280 } 281 282 inline void build_t() { 283 std::sort(imp + 1, imp + cnt + 1, cmp); 284 cnt = std::unique(imp + 1, imp + cnt + 1) - imp - 1; 285 286 stk[top = 1] = imp[1]; 287 for(int i = 2; i <= cnt; i++) { 288 int x = imp[i], y = lca(x, stk[top]); 289 work(x); work(y); 290 while(top > 1 && pos2[y] <= pos2[stk[top - 1]]) { 291 ADD(stk[top - 1], stk[top]); 292 top--; 293 } 294 if(y != stk[top]) { 295 ADD(y, stk[top]); 296 stk[top] = y; 297 } 298 stk[++top] = x; 299 } 300 while(top > 1) { 301 ADD(stk[top - 1], stk[top]); 302 top--; 303 } 304 return; 305 } 306 307 inline void solve() { 308 309 for(int i = 1; i <= n; i++) { 310 vx[i].clear(); 311 } 312 313 for(int i = 1; i <= m; i++) { 314 vx[node[i].w].push_back(i); 315 } 316 317 for(int rt = 1; rt <= n; rt++) { 318 if(!vx[rt].size()) continue; 319 ++Time; 320 TP = cnt = 0; 321 work(rt); 322 imp[++cnt] = rt; 323 for(int j = 0; j < (int)vx[rt].size(); j++) { 324 int i = vx[rt][j]; 325 int x = node[i].x, y = node[i].y, z = node[i].w; 326 work(x); work(y); 327 imp[++cnt] = x; 328 imp[++cnt] = y; 329 if(x != z) { 330 fr[n + i * 2 - 1] = y; 331 d[n + i * 2 - 1] = d[y] + d[x] + node[i].val - node[i].z; 332 v[x].push_back(n + i * 2 - 1); 333 } 334 if(y != z) { 335 fr[n + i * 2] = x; 336 d[n + i * 2] = d[x] + d[y] + node[i].val - node[i].z; 337 v[y].push_back(n + i * 2); 338 } 339 } 340 build_t(); 341 dfs(rt, 0); 342 } 343 return; 344 } 345 } 346 347 inline void solve() { 348 Ans = -INF; 349 scanf("%d", &n); 350 for(int i = 1; i <= n; i++) { 351 fr[i] = i; 352 } 353 LL z; 354 for(int i = 1, x, y; i < n; i++) { 355 scanf("%d%d%lld", &x, &y, &z); 356 add(x, y, z); 357 add(y, x, z); 358 } 359 360 DFS(1, 0); 361 prework(); 362 363 scanf("%d", &m); 364 for(int i = 1, x, y; i <= m; i++) { 365 scanf("%d%d%lld", &x, &y, &z); 366 node[i].x = x; 367 node[i].y = y; 368 node[i].z = z; 369 node[i].val = dis(x, y) - z; 370 node[i].w = lca(x, y); 371 } 372 373 Sone::solve(); 374 Stwo::solve(); 375 376 if(Ans < (LL)-1e18) puts("F"); 377 else printf("%lld ", Ans); 378 return; 379 } 380 381 int main() { 382 383 for(int i = 2; i < N; i++) pw[i] = pw[i >> 1] + 1; 384 385 int T; 386 scanf("%d", &T); 387 while(T--) { 388 solve(); 389 if(T) clear(); 390 } 391 return 0; 392 }
这什么鬼题啊...搞了4天，结果写完啥都没学到......