A. The Fair Nut and Elevator
Solved.
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 110 5 int n, a[N]; 6 7 int main() 8 { 9 while (scanf("%d", &n) != EOF) 10 { 11 for (int i = 1; i <= n; ++i) scanf("%d", a + i); 12 int res = 1e9; 13 for (int x = 1, tmp = 0; x <= 100; ++x, res = min(res, tmp), tmp = 0) for (int i = 1; i <= n; ++i) if (a[i]) 14 { 15 tmp += a[i] * (abs(x - i) + abs(i - 1) + abs(x - 1) + abs(x - 1) + abs(i - 1) + abs(x - i)); 16 } 17 printf("%d ", res); 18 } 19 return 0; 20 }
B. Kvass and the Fair Nut
Upsolved.
题意：
刚开始题意读错，过了pretest就没管了
有$n桶啤酒，要从中取出s升，求取出后剩余的量最少的那桶啤酒最大$
思路：
二分或者直接算都行
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 1010 6 int n; 7 ll s, v[N]; 8 9 bool check(ll x) 10 { 11 ll tot = 0; 12 for (int i = 1; i <= n; ++i) 13 { 14 if (v[i] < x) return false; 15 tot += v[i] - x; 16 } 17 return tot >= s; 18 } 19 20 int main() 21 { 22 while (scanf("%d%lld", &n, &s) != EOF) 23 { 24 for (int i = 1; i <= n; ++i) scanf("%lld", v + i); 25 ll l = 0, r = 1e9, res = -1; 26 while (l <= r) 27 { 28 ll mid = (l + r) >> 1; 29 if (check(mid)) 30 { 31 res = mid; 32 l = mid + 1; 33 } 34 else 35 r = mid - 1; 36 } 37 printf("%lld ", res); 38 } 39 return 0; 40 }
C. The Fair Nut and String
Solved.
题意：
有一串字符串，求有多少个字符串，是abababab形式的。
思路：
字符不是‘a’ 也不是‘b’ 的话是没用的，然后如果两个'a'之间有多个'b'也没用
缩短字符串后
必然是aaabaaababab 这样的形式的
那么像统计DAG那样统计就没了
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 100010 6 const ll MOD = (ll)1e9 + 7; 7 char s[N], t[N]; 8 9 int main() 10 { 11 while (scanf("%s", s + 1) != EOF) 12 { 13 int n = 0; 14 for (int i = 1, len = strlen(s + 1); i <= len; ++i) if (s[i] == 'a' || s[i] == 'b') 15 { 16 if (s[i] == 'a') t[++n] = 'a'; 17 else if (s[i] == 'b' && t[n] == 'a') t[++n] = 'b'; 18 } 19 if (n == 0) 20 { 21 puts("0"); 22 continue; 23 } 24 ll res = 1, pre = 1, cnt = 1; 25 for (int i = 2; i <= n; ++i) 26 { 27 if (t[i] == 'b') 28 { 29 pre = (pre * (cnt + 1)) % MOD; 30 cnt = 0; 31 } 32 else 33 { 34 ++cnt; 35 res = (res + pre) % MOD; 36 } 37 } 38 printf("%lld ", res); 39 } 40 return 0; 41 }
D. The Fair Nut and the Best Path
Upsolved.
题意：
在一棵树上，每个点是加油站，最多加油$w_i，然后每条边是路，耗费油v_i$
$定义f(u, v)为u->v的简单路径上每个点都加满油，假设油箱容量无限，最后剩下的油量$
如果其中某条路上油耗尽了，那么这条路是不可行的
思路：
我们把点权视为正直，边权视为负值
然后就是求任意两点之间的最大权值和
不需要考虑不合法的路径，因为如果存在不合法的路劲，
那么肯定存在另一条合法的路径使得答案比它更优。
令$f[u] 表示到达u的子树中某点的最大权， 这是纵向路径$
再考虑横向路径即可
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 300010 6 #define pii pair <int, int> 7 int n, w[N]; 8 vector <pii> G[N]; 9 ll res, f[N]; 10 11 void DFS(int u, int fa) 12 { 13 f[u] = w[u]; 14 ll Max[2] = {0, 0}; 15 for (auto it : G[u]) 16 { 17 int v = it.first; 18 if (v == fa) continue; 19 DFS(v, u); 20 ll cost = it.second; 21 f[u] = max(f[u], f[v] - cost + w[u]); 22 ll tmp = f[v] - cost; 23 if (tmp > Max[0]) 24 { 25 Max[1] = Max[0]; 26 Max[0] = tmp; 27 } 28 else if (tmp > Max[1]) 29 Max[1] = tmp; 30 } 31 res = max(res, max(f[u], Max[0] + Max[1] + w[u])); 32 } 33 34 int main() 35 { 36 while (scanf("%d", &n) != EOF) 37 { 38 for (int i = 1; i <= n; ++i) scanf("%d", w + i); 39 for (int i = 1; i <= n; ++i) G[i].clear(); 40 for (int i = 1, u, v, w; i < n; ++i) 41 { 42 scanf("%d%d%d", &u, &v, &w); 43 G[u].emplace_back(v, w); 44 G[v].emplace_back(u, w); 45 } 46 res = 0; 47 DFS(1, 1); 48 printf("%lld ", res); 49 } 50 return 0; 51 }
E. The Fair Nut and Strings
Upsolved.
题意：
一共有$k个长度为n的字符串，他们的范围是[s, t] 之间，按字典序排序$
求这些字符串(构造k个满足字典序要求的字符串)最多有多少个前缀
思路：
相当于给出一棵二叉字典树，给出左右界，叶子节点不超过$k个$
求最多节点个数
能扩展就扩展，贪心一下即可
注意特判$k = 1$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 500010 6 int n, k; 7 char s[N], t[N]; 8 9 int main() 10 { 11 while (scanf("%d%d", &n, &k) != EOF) 12 { 13 scanf("%s%s", s + 1, t + 1); 14 if (k == 1) 15 { 16 printf("%d ", n); 17 continue; 18 } 19 int flag = 0; 20 ll cur = 0, res = 0; k -= 2; 21 for (int i = 1; i <= n; ++i) 22 { 23 if (flag == 0 && s[i] != t[i]) 24 flag = 1; 25 else if (flag == 1) 26 { 27 ll extend = cur; 28 if (s[i] == 'a') ++extend; 29 if (t[i] == 'b') ++extend; 30 if (extend <= k) 31 { 32 k -= extend; 33 cur += extend; 34 } 35 else 36 { 37 cur += k; 38 k = 0; 39 } 40 } 41 res += cur + 1 + flag; 42 } 43 printf("%lld ", res); 44 } 45 return 0; 46 }
F. Max Mex
Upsolved.
题意：
一棵树上，两种操作
$交换两点的p[]值$
$询问MEX(v(l)), l表示某挑简单路径上权值的集合$
思路：
我们考虑将点按权值排序
我们考虑$1 -> i 和 i + 1 -> j$
$这两段的合并$
如果可以合并，那么答案的下限就是$j$
$树上的路径有三种类型$
$1、一个点$
$2、一条链$
$3、两条链$
那分情况讨论，一共只有6种情况
$但是这样太麻烦了$
$我们考虑这三种状态都可以用两条链的情况来表示$
$那么合并的时候，一条路径可以用两个点来表示$
$那么一条路径如果包含另外一条路径的两个端点$
$那么这条路径就包含另外一条路径$
$也就是说我们只需要分两次合并另外一条路径的两点即可$
$这样就只有一种情况$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 200010 5 int n, q, p[N]; 6 vector <int> G[N]; 7 8 int in[N], out[N]; 9 namespace ST 10 { 11 int rmq[N << 1]; 12 int mm[N << 1]; 13 int dp[N << 1][20]; 14 int F[N << 1], P[N]; 15 int cnt, cnt2; 16 void init(int n) 17 { 18 mm[0] = -1; 19 for (int i = 1; i <= n; ++i) 20 { 21 mm[i] = ((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1]; 22 dp[i][0] = i; 23 } 24 for (int j = 1; j <= mm[n]; ++j) 25 for (int i = 1; i + (1 << j) - 1 <= n; ++i) 26 { 27 dp[i][j] = rmq[dp[i][j - 1]] < rmq[dp[i + (1 << (j - 1))][j - 1]] ? 28 dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1]; 29 } 30 } 31 void DFS(int u, int pre, int dep) 32 { 33 F[++cnt] = u; 34 rmq[cnt] = dep; 35 P[u] = cnt; 36 in[u] = ++cnt2; 37 for (auto v : G[u]) if (v != pre) 38 { 39 DFS(v, u, dep + 1); 40 F[++cnt] = u; 41 rmq[cnt] = dep; 42 } 43 out[u] = cnt2; 44 } 45 void init(int root, int node_num) 46 { 47 cnt = 0; cnt2 = 0; 48 DFS(root, root, 0); 49 init(2 * node_num - 1); 50 } 51 int query(int a, int b) 52 { 53 a = P[a], b = P[b]; 54 if (a > b) swap(a, b); 55 int k = mm[b - a + 1]; 56 return F[rmq[dp[a][k]] <= rmq[dp[b - (1 << k) + 1][k]] ? 57 dp[a][k] : dp[b - (1 << k) + 1][k]]; 58 } 59 } 60 61 namespace SEG 62 { 63 struct node 64 { 65 int u, v, p; 66 void init() { u = v = p = 0; } 67 node () {} 68 node (int u, int v, int p) : u(u), v(v), p(p) {} 69 }a[N << 2], res; 70 int ans; 71 void init() { memset(a, 0, sizeof a); } 72 bool anc(int x, int y) 73 { 74 return in[x] <= in[y] && out[x] >= out[y]; 75 } 76 node check(node a, int y) 77 { 78 if (a.u == -1 || y == -1) return node(-1, -1, -1); 79 if (!y) return a; 80 if (!a.u) return node(y, y, y); 81 if (!anc(a.u, a.v)) swap(a.u, a.v); 82 if (anc(a.u, a.v)) 83 { 84 if (anc(a.u, y)) 85 { 86 int p = ST::query(y, a.v); 87 if (p == y) 88 return a; 89 else if (p == a.u) 90 return node(a.v, y, a.u); 91 else if (p == a.v) 92 return node(a.u, y, a.u); 93 else 94 return node(-1, -1, -1); 95 } 96 else if (anc(y, a.u)) 97 return node(y, a.v, y); 98 else 99 { 100 int p = ST::query(a.v, y); 101 return node(a.v, y, p); 102 } 103 } 104 else if (anc(a.p, y)) 105 { 106 if (anc(a.u, y)) 107 return node(y, a.v, a.p); 108 else if (anc(a.v, y)) 109 return node(a.u, y, a.p); 110 else if (anc(y, a.u) || anc(y, a.v)) 111 return a; 112 else 113 return node(-1, -1, -1); 114 } 115 else 116 return node(-1, -1, -1); 117 } 118 node merge(node a, node b) 119 { 120 a = check(a, b.u); 121 a = check(a, b.v); 122 return a; 123 } 124 void update(int id, int l, int r, int pos, int v) 125 { 126 if (l == r) 127 { 128 a[id] = node(v, v, v); 129 return; 130 } 131 int mid = (l + r) >> 1; 132 if (pos <= mid) update(id << 1, l, mid, pos, v); 133 else update(id << 1 | 1, mid + 1, r, pos, v); 134 a[id] = merge(a[id << 1], a[id << 1 | 1]); 135 // printf("%d %d %d %d %d %d ", l, r, a[id].t, a[id].u, a[id].v, a[id].p); 136 // printf("%d %d %d %d %d %d ", l, mid, a[id << 1].t, a[id << 1].u, a[id << 1].v, a[id << 1].p); 137 // printf("%d %d %d %d %d %d ", mid + 1, r, a[id << 1 | 1].t, a[id << 1 | 1].u, a[id << 1 | 1].v, a[id << 1 | 1].p); 138 // puts("*************************************"); 139 } 140 bool query(int id, int l, int r) 141 { 142 node tmp = merge(res, a[id]); 143 // printf("bug %d %d %d %d %d ", l, r, a[id].u, a[id].v, a[id].p); 144 // printf("bug %d %d %d %d %d ", l, r, res.u, res.v, res.p); 145 // printf("bug %d %d %d %d %d ", l, r, tmp.u, tmp.v, tmp.p); 146 // puts("**********************"); 147 if (tmp.u != -1) 148 { 149 res = tmp; 150 ans = r; 151 return 1; 152 } 153 if (l == r) return 0; 154 int mid = (l + r) >> 1; 155 if (query(id << 1, l, mid)) 156 query(id << 1 | 1, mid + 1, r); 157 return 0; 158 } 159 } 160 161 int main() 162 { 163 while (scanf("%d", &n) != EOF) 164 { 165 for (int i = 1; i <= n; ++i) G[i].clear(); 166 for (int i = 1; i <= n; ++i) scanf("%d", p + i), p[i] += 1; 167 for (int v = 2, u; v <= n; ++v) 168 { 169 scanf("%d", &u); 170 G[u].push_back(v); 171 G[v].push_back(u); 172 } 173 ST::init(1, n); 174 SEG::init(); 175 for (int i = 1; i <= n; ++i) 176 { 177 //printf("%d %d ", p[i], i); 178 SEG::update(1, 1, n, p[i], i); 179 } 180 scanf("%d", &q); 181 for (int i = 1, op, x, y; i <= q; ++i) 182 { 183 scanf("%d", &op); 184 if (op == 1) 185 { 186 scanf("%d%d", &x, &y); 187 swap(p[x], p[y]); 188 SEG::update(1, 1, n, p[x], x); 189 SEG::update(1, 1, n, p[y], y); 190 } 191 else 192 { 193 //for (int i = 1; i <= n; ++i) printf("%d%c", p[i], " "[i == n]); 194 SEG::res.init(); SEG::ans = 1; 195 SEG::query(1, 1, n); 196 printf("%d ", SEG::ans); 197 } 198 } 199 } 200 return 0; 201 }