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 
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);
		else
			printf("Case %d: %.6f\n",++cas,dp[1].Ci/(1.0-dp[1].Ai));
	}
	return 0;
}