标签:size ORC str codeforce 代码 efi include har mes
[题目链接]
https://codeforces.com/contest/715/problem/C
[算法]
考虑点分治
一条路径(x , y)合法当且仅当 : d(x) * 10 ^ dep(x) + d(y) = 0(mod m) , 其中d(u)表示u到分治重心路径上数字拼接起来所形成的数
统计答案时 , 我们只需维护一个map , 维护10 ^ -dep(u) * d(u) (mod m)
然后计算每个点的贡献即可
时间复杂度 : O(NlogN ^ 2)
[代码]
#include<bits/stdc++.h> using namespace std; #define N 100010 typedef long long ll; typedef long double ld; typedef unsigned long long ull; struct edge { int to , w , nxt; } e[N << 1]; int n , m , tot , len , root; ll ans; int pw[N] , head[N] , size[N] , weight[N] , D[N] , depth[N]; bool visited[N]; map<int , int> mp; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘; x *= f; } inline void addedge(int u , int v , int w) { ++tot; e[tot] = (edge){v , w , head[u]}; head[u] = tot; } inline void getroot(int u , int par , int total) { size[u] = 1; weight[u] = 0; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v == par || visited[v]) continue; getroot(v , u , total); size[u] += size[v]; chkmax(weight[u] , size[v]); } chkmax(weight[u] , total - size[u]); if (weight[u] < weight[root]) root = u; } inline void exgcd(int a , int b , int &x , int &y) { if (b == 0) { x = 1; y = 0; } else { exgcd(b , a % b , y , x); y -= a / b * x; } } inline int inv(int a) { int x , y; exgcd(a , m , x , y); return (x % m + m) % m; } inline void dfs(int u , int par , int d1 , int d2) { if (depth[u] > 0) ++mp[(1ll * d2 % m * inv(pw[depth[u]] % m)) % m]; D[++len] = d1; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to , w = e[i].w; if (v == par || visited[v]) continue; depth[v] = depth[u] + 1; dfs(v , u , (1ll * w * pw[depth[v] - 1] % m + d1) % m , (10ll * d2 % m + w) % m); } } inline ll calc(int u , int d) { mp.clear(); len = 0; if (!d) dfs(u , -1 , 0 , 0); else dfs(u , -1 , d % m , d % m); ll res = 0; for (int i = 1; i <= len; ++i) { int goal = ((m - D[i]) % m + m) % m; res += (ll)mp[goal]; if (!d && !D[i]) ++res; } return res; } inline void work(int u) { visited[u] = true; depth[u] = 0; ans += calc(u , 0); for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to , w = e[i].w; if (visited[v]) continue; depth[v] = 1; ans -= calc(v , w); } for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (visited[v]) continue; root = 0; getroot(v , u , size[v]); work(root); } } int main() { read(n); read(m); pw[0] = 1; for (int i = 1; i <= n; ++i) pw[i] = 1ll * pw[i - 1] * 10 % m; for (int i = 1; i < n; ++i) { int u , v , w; read(u); read(v); read(w); ++u; ++v; addedge(u , v , w); addedge(v , u , w); } weight[0] = n; root = 0; getroot(1 , 0 , n); work(root); ans -= n; printf("%I64d\n" , ans); return 0; }
标签:size ORC str codeforce 代码 efi include har mes
原文地址:https://www.cnblogs.com/evenbao/p/10544912.html