hdu 4035 Maze

第一次做概率DP，找了篇大神的借鉴一下

个人感觉，根据题意，每个数据节点均要父节点与子节点的数据联合计算，因此可以将问题化简为系数表达式

#include 
#include 
#include 
#include 
using namespace std;
// hdu 4035 
/*
	概率dp求期望
	e[i]表示从i走出迷宫需要的边数期望
	叶子节点：
		e[i] = bk[i]*e[1] + fd[i]*0 + (1-bk-fd)*(1+e[mother[i]]);
	非叶子节点：
		e[i] = bk[i]*e[1] + fd[i]*0 + (1-bk-fd)/m*(1+e[mother[i]]+Σ(e[child[i]]));
*/
const int MAXN = 10005;
const double _inf = 1e-10;
int n;
vector mp[MAXN];
double bk[MAXN], fd[MAXN], mother[MAXN];
double A[MAXN], B[MAXN], C[MAXN];
bool dfs(int i, int mo)
{
	if (mp[i].size() == 1 && mo) // 叶子节点&非节点1
	{
		A[i] = bk[i];
		B[i] = 1 - bk[i] - fd[i];
		C[i] = 1 - bk[i] - fd[i];
		return true;
	}
	// 节点1 或 非叶子节点
	A[i] = bk[i];
	B[i] = (1 - bk[i] - fd[i])/mp[i].size();
	C[i] = 1 - bk[i] - fd[i];
	double tmp = 0.0;
	for (int j = 0; j< mp[i].size(); ++j)
	{
		if ( mp[i][j] == mo ) continue;
		if (!dfs(mp[i][j], i)) return false;
		A[i] += A[mp[i][j]] * B[i];
		C[i] += C[mp[i][j]] * B[i];
		tmp += B[mp[i][j]] * B[i];
	}
	if (fabs(1-tmp) < _inf) return false;
	A[i] /= 1 - tmp;
	B[i] /= 1 - tmp;
	C[i] /= 1 - tmp;
	return true;
}
int main()   
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
	int t, cs = 0;
	int i, j, k;
	scanf("%d", &t);
	while (t--)
	{
		printf("Case %d: ", ++cs);
		scanf("%d", &n);
		for (i = 1; i<= n; ++i) mp[i].clear();
		for (i = 1; i< n; ++i) scanf("%d%d", &j, &k), 
			mp[j].push_back(k), mp[k].push_back(j);
		for ( i = 1; i<= n; ++i)
		{
			scanf("%lf%lf", &bk[i], &fd[i]);
			bk[i] /= 100.0;
			fd[i] /= 100.0;
		}
		if (dfs(1, 0) && fabs(1-A[1]) > _inf)
			printf("%lf\n", C[1]/(1-A[1]));
		else
			printf("impossible\n");
	}
	return 0;
}