题目及数据：
https://www.lanzous.com/i1xdr8h
心路历程
预计分数：$100+ 100 +60 = 260$
实际得分：$100 + 100 + 0 = 200$
我这次检查了三遍绝对没算错！！！
上来看T1，咦？我好像做过这题在仙人掌上的版本。。树上更简单吧。。写+拍 1h，期间拍出了暴力的两个bug。。。
看T2，好难啊，不会做啊 qwq。。然后开始强行套算法。恩。把单调队列套进去发现是可行的。。而且好像还可以无视题目的一些性质。。嘿嘿嘿这题有加强版了
T3更神仙。。想了一个并查集优化暴力的$n^2$算法，然后写挂了。。。自己造的输出跑的贼快，一上评测机就死循环。。
因为昨天晚上打cf特别困，而且自我感觉$260$应该不低了，就睡了一个半小时。。。。。。。。。
T1
听大佬们说是原题啊Orz
我居然不知道 我好像是做过然后忘的一干二净。。
很显然，把路径拆开，判断lca之间的关系即可
#include<cstdio> #include<cstdlib> #include<algorithm> #include<vector> #include<cstring> #define getchar() ((p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++) using namespace std; const int MAXN = 1e5 + 10; char buf[(1 << 21)], *p1 = buf, *p2 = buf; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int T, N, Q, fa[MAXN], dep[MAXN], top[MAXN], son[MAXN], siz[MAXN]; vector<int> v[MAXN]; void init() { for(int i = 1; i <= N; i++) v[i].clear(); memset(top, 0, sizeof(top)); memset(son, 0, sizeof(son)); } void dfs1(int x, int _fa) { siz[x] = 1; fa[x] = _fa; for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(to == _fa) continue; dep[to] = dep[x] + 1; dfs1(to, x); siz[x] += siz[to]; if(siz[to] > siz[son[x]]) son[x] = to; } } void dfs2(int x, int topf) { top[x] = topf; if(!son[x]) return ; dfs2(son[x], topf); for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(top[to]) continue; dfs2(to, to); } } int LCA(int x, int y) { while(top[x] != top[y]) { if(dep[top[x]] < dep[top[y]]) swap(x, y); x = fa[top[x]]; } if(dep[x] > dep[y]) swap(x, y); return x; } bool check(int a, int x, int y) { if(dep[x] < dep[y]) swap(x, y); if(LCA(a, x) == a && LCA(a, y) == y) return 1; else return 0; } int main() { freopen("railway.in", "r", stdin); freopen("railway.out", "w", stdout); T = read(); while(T--) { N = read(); Q = read(); init(); for(int i = 1; i <= N - 1; i++) { int x = read(), y = read(); v[x].push_back(y); v[y].push_back(x); } dep[1] = 1; dfs1(1, 0); dfs2(1, 1); while(Q--) { int xx1 = read(), yy1 = read(), xx2 = read(), yy2 = read(); int lca1 = LCA(xx1, yy1), lca2 = LCA(xx2, yy2); if(check(lca1, xx2, lca2) || check(lca1, yy2, lca2) || check(lca2, xx1, lca1) || check(lca2, yy1, lca1)) puts("YES"); else puts("NO"); } } return 0; }
T2
式子化简完了之后应该是求
$$sum_{i = 1}^n sum_{j = 1}^i mx - mn$$
其中
$mx = max(a[j], a[j+1], dots a[i])$
$mn = min(a[j], a[j+1], dots a[i])$
单调栈维护即可。。。。
#include<cstdio> #include<cstdlib> #include<algorithm> #include<iostream> #define LL long long using namespace std; const LL MAXN = 1e5 + 10; inline LL read() { char c = getchar(); LL x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } LL T, N; LL a[MAXN], q[MAXN]; LL solve() { LL h = 1, t = 0, ans = 0, sum = 0; for(LL i = 1; i <= N; i++) { while(h <= t && a[i] > a[q[t]]) sum -= a[q[t]] * (q[t] - q[t - 1]), t--; q[++t] = i; ans += a[i] * (q[t] - q[t - 1]) + sum; sum += a[i] * (q[t] - q[t - 1]); } return ans; } int main() { freopen("count.in", "r", stdin); freopen("count.out", "w", stdout); T = read(); while(T--) { N = read(); for(LL i = 1; i <= N; i++) a[i] = read(); LL ans = solve(); for(LL i = 1; i <= N; i++) a[i] = -a[i]; LL ans2 = solve(); cout << ans + ans2 << endl; } return 0; }
T3
不明白出这种题有什么意义。。。
dsu on tree， set，这都是noip知识点？？？？
而且标算好像就是在想尽各种方法优化暴力。。。
考虑直接算一条边的贡献
首先考虑暴力怎么算
对于一条边来说，如果子树内有连续的区间，比如$[1, 2, 3, 4]$那么在统计他们的答案时显然不会经过该边，他们对答案的贡献为$frac{n(n -1)}{2} $
用总贡献减去即可。
然后可以分成子树内和子树外讨论，子树内用并查集维护，子树外用set维护并查集的补集
做完了。。。
代码里面的solve函数是抄的标算，，，实在写不出来啊qwq
#include<bits/stdc++.h> #define LL long long using namespace std; const int MAXN = 1e5 + 10; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, dep[MAXN], siz[MAXN], son[MAXN], dsu[MAXN], Son, vis[MAXN], ds[MAXN]; vector<int> v[MAXN]; set<int> s; LL AnsOut, AnsIn, ans; void dfs(int x, int _fa) { siz[x] = 1; for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(to == _fa) continue; dfs(to, x); siz[x] += siz[to]; if(siz[to] > siz[son[x]]) son[x] = to; } } LL calc(LL x) { return x * (x - 1) / 2; } void Clear() { s.clear(); AnsOut = calc(N); AnsIn = 0; s.insert(0); s.insert(N + 1); } int find(int x) { return dsu[x] == x ? dsu[x] : dsu[x] = find(dsu[x]); } void solve(int x) { s.insert(x); set<int>::iterator s1, s2, it; s1 = s2 = it = s.find(x); s1--; s2++; AnsOut -= calc((*s2) - (*s1) - 1); AnsOut += calc((*s2) - (*it) - 1) + calc((*it) - (*s1) - 1); vis[x] = 1; if(vis[x - 1]) { int fx = find(x - 1), fy = find(x); AnsIn += ds[fx] * ds[fy]; dsu[fx] = fy; ds[fy] += ds[fx]; } if(vis[x + 1]) { int fx = find(x + 1), fy = find(x); AnsIn += ds[fx] * ds[fy]; dsu[fx] = fy; ds[fy] += ds[fx]; } } void Add(int x, int fa) { solve(x); for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(to == fa || to == Son) continue; Add(to, x); } } void Delet(int x, int fa) { vis[x] = 0; ds[x] = 1; dsu[x] = x; for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(to == fa) continue; Delet(to, x); } } void dfs2(int x, int fa, int opt) { for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(to == fa || (to == son[x])) continue; dfs2(to, x, 0); } if(son[x]) dfs2(son[x], x, 1); Son = son[x]; Add(x, fa); ans += calc(N) - AnsIn - AnsOut; if(opt == 0) Delet(x, fa), Clear(), Son = 0; } main() { N = read(); for(int i = 1; i <= N - 1; i++) { int x = read(), y = read(); v[x].push_back(y); v[y].push_back(x); } for(int i = 1; i <= N; i++) dsu[i] = i, ds[i] = 1; dep[1] = 1; dfs(1, 0); Clear(); dfs2(1, 0, 0); cout << ans; return 0; } /* 10 1 9 9 7 9 5 5 3 9 4 4 8 1 10 1 2 3 6 4 1 4 1 3 2 4 */