HDU 4035 Maze 概率dp+树形dp

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题解：点击打开链接

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
using namespace std;
const double eps = 1e-9;
const int N = 10010;
vector<int> G[N];
int n;
double k[N], e[N], dp[N];
double A[N], B[N], C[N];
bool dfs(int u, int fa){
	int m = G[u].size();
	A[u] = k[u];
	B[u] = (1 - k[u] - e[u])/m;
	C[u] = 1 - k[u] - e[u];
	double tmp = (1 - k[u] - e[u]) / (double)m, self = 0;
	for(int i = G[u].size()-1; i >= 0; i--){
		int v = G[u][i]; if(v == fa)continue;
		if(!dfs(v, u)) return false;
		A[u] +=  tmp * A[v];
		self += tmp * B[v];
		C[u] += tmp * C[v];
	}
	if(fabs(self - 1.0) < eps) return false;
	A[u] /= 1-self;
	B[u] /= 1-self;
	C[u] /= 1-self;
	return true;
}

void input(){
	cin>>n;
	for(int i = 1, u, v; i < n; i++){
		scanf("%d %d", &u, &v);
		G[u].push_back(v); G[v].push_back(u);
	}
	for(int i = 1; i <= n; i++) {
		scanf("%lf %lf", &k[i], &e[i]);
		k[i] /= 100.0;
		e[i] /= 100.0;
	}
}
int main() {
	int T, Cas = 1; cin>>T;
    while(T--){
    	input();
    	bool ok = dfs(1, -1);

    	printf("Case %d: ", Cas++);
    	if(!ok || fabs(1.0-A[1]) < eps)
    		puts("impossible");
    	else
    		printf("%.6f\n", C[1] / (1.0-A[1]));

    	for(int i = 1; i <= n; i++) G[i].clear();
    }
    return 0;
}

HDU 4035 Maze 概率dp+树形dp

标签：blog   http   io   os   ar   for   sp   2014   on   

原文地址：http://blog.csdn.net/qq574857122/article/details/40344319