BZOJ 3566 SHOI 2014 概率充电器 概率DP

题目大意：给出一棵树，每个节点都有一个充电概率，每一条边有一个导电概率，求期望有多少个点充电。

思路：写不出题解，粘一个详细的地址：http://wyfcyx.is-programmer.com/posts/74623.html

CODE：

#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <cstring>#include <iomanip>#include <iostream>#include <algorithm>#define EPS 1e-8#define MAX 500010using namespace std;int points;double src[MAX];int head[MAX],total;int _next[MAX << 1],aim[MAX << 1];double length[MAX << 1];int father[MAX];double f[MAX][2],p[MAX];inline void Add(int x,int y,double len){_next[++total] = head[x];aim[total] = y;length[total] = len;head[x] = total;}void DFS1(int x,int last){f[x][0] = 1.0 - src[x];father[x] = last;for(int i = head[x]; i; i = _next[i]) {if(aim[i] == last)continue;p[aim[i]] = length[i];DFS1(aim[i],x);f[x][0] *= (f[aim[i]][0] + (1 - f[aim[i]][0]) * (1 - length[i]));}}void DFS2(int x,int last){for(int i = head[x]; i; i = _next[i]) {if(aim[i] == last)continue;double t = f[aim[i]][0] + (1 - f[aim[i]][0]) * (1 - length[i]);double _p = t < EPS ? 0:f[x][1] * f[x][0] / t;f[aim[i]][1] = _p + (1 - _p) * (1 - length[i]);DFS2(aim[i],x);}}int main(){cin >> points;for(int x,y,z,i = 1; i < points; ++i) {scanf("%d%d%d",&x,&y,&z);Add(x,y,(double)z / 100.0),Add(y,x,(double)z / 100.0);}for(int i = 1; i <= points; ++i) {scanf("%lf",&src[i]);src[i] /= 100.0;}DFS1(1,0);f[1][1] = 1;DFS2(1,0);double ans = .0;for(int i = 1; i <= points; ++i)ans += (1 - f[i][0] * f[i][1]);cout << fixed << setprecision(6) << ans << endl;return 0;}