HDU 4035 Maze 機率dp+樹形dp

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題解：點擊開啟連結

#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 機率dp+樹形dp