[概率期望][矩阵树][高斯消元]luogu P3317 重建

https://www.luogu.org/problem/P3317

分析

题目要求$sum_{T} (prod_{ein T} p_e prod_{e otin T}(1-p_e) )$

乍一看很像矩阵树，但是又不是

我们化一下柿子：

$sum_{T} (prod_{ein T} p_e frac{prod_{e}(1-p_e)}{prod_{ein T}(1-p_e)} )$

$prod_{e}(1-p_e)sum_{T} ( frac{prod_{ein T} p_e}{prod_{ein T}(1-p_e)} )$

套矩阵树模板就可以了

#include <iostream>
#include <cstdio>
using namespace std;
typedef double db;
const db eps=1e-8;
const int N=60;
int n;
db g[N][N],k=1.0,ans=1.0;

void Gauss(int n) {
    int mx;db t;
    for (int i=1;i<=n;i++) {
        mx=i;
        for (int j=i+1;j<=n;j++) if (g[mx][i]<g[j][i]) mx=j;
        if (mx!=i) swap(g[mx],g[i]);
        if (g[i][i]==0) {
            ans=0;
            return;
        }
        for (int j=i+1;j<=n;j++) {
            t=g[j][i]/g[i][i];
            for (int k=i;k<=n;k++)
                g[j][k]-=g[i][k]*t;
        }
        ans*=g[i][i];
    }
    if (ans<0) ans=-ans;
}

int main() {
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) {
            scanf("%lf",&g[i][j]);
            if (g[i][j]==1) g[i][j]=1.0-eps;
            if (i<j) k*=1.0-g[i][j];
            g[i][j]=g[i][j]/(1.0-g[i][j]);
        }
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
            if (j!=i) g[i][i]+=g[i][j],g[i][j]=-g[i][j];
    Gauss(n-1);
    printf("%.9lf",ans*k);
}