hdu 4035 概率dp

很好的一个题目

具体思路网上说的很多

我是参考http://www.cnblogs.com/kuangbin/archive/2012/10/03/2711108.html这个的，感觉写的很浅显易懂

存个代码

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 10000+10;const double eps = 1e-9;int T,n;int head[maxn],_cnt,vist[maxn],degree[maxn];struct Edge{int to,next;}edge[maxn<<1];inline void init(){for(int i = 0; i <= n; i++){head[i] = -1;vist[i] = 0;degree[i] = 0;}_cnt = 0;}inline void ADD(int u,int v){edge[_cnt].to = v;edge[_cnt].next = head[u];head[u] = _cnt++;}struct Info{double ki,ei;}nxt[maxn];struct DP{double Ai,Bi,Ci;}dp[maxn];inline void dfs(int now){vist[now] = 1;if(head[now] == -1){dp[now].Ai = nxt[now].ki;dp[now].Bi = 1.0-nxt[now].ei-nxt[now].ki;dp[now].Ci = 1.0-nxt[now].ei-nxt[now].ki;return;}double sum_Aj,sum_Bj,sum_Cj;sum_Aj = sum_Bj = sum_Cj = 0.0;for(int i = head[now]; ~i; i = edge[i].next){int v = edge[i].to;if(!vist[v]){dfs(v);sum_Aj += dp[v].Ai;sum_Bj += dp[v].Bi;sum_Cj += dp[v].Ci;}}double temp = 1.0-((1-nxt[now].ei-nxt[now].ki)*sum_Bj)/degree[now];dp[now].Ai = (nxt[now].ki+(1-nxt[now].ei-nxt[now].ki)*sum_Aj/degree[now])/temp;dp[now].Bi = ((1-nxt[now].ei-nxt[now].ki)/degree[now])/temp;dp[now].Ci = (1-nxt[now].ei-nxt[now].ki+(1-nxt[now].ei-nxt[now].ki)*sum_Cj/degree[now])/temp;}int main(){int cas = 0;int u,v;scanf("%d",&T);while(T--){scanf("%d",&n);init();for(int i = 1; i < n; i++){scanf("%d %d",&u,&v);degree[u]++;degree[v]++;ADD(u,v);ADD(v,u);}for(int i = 1; i <= n; i++){scanf("%lf %lf",&nxt[i].ki,&nxt[i].ei);nxt[i].ki /= 100.0;nxt[i].ei /= 100.0;}dfs(1);if(fabs(1-dp[1].Ai) < eps)printf("Case %d: impossible\n",++cas);elseprintf("Case %d: %.6f\n",++cas,dp[1].Ci/(1.0-dp[1].Ai));}return 0;}