[luoguP2973]Driving Out the Piggies G

题意

一个无向图，节点 \(1\) 有一个炸弹，在每个单位时间内，有 \(\dfrac{p}{q}\) 的概率在这个节点炸掉，有 \(1-\dfrac{p}{q}\) 的概率随机选择一条出去的路到其他的节点上，问最终炸弹在每个节点上爆炸的概率。

sol

由于最终炸弹在爆炸后就无法再移动了，因此炸弹到达每个点的次数的期望 \(E \cdot \dfrac{p}{q}\) 即为爆炸的概率。
因此我们可以列出方程

\[f_u=[u=1] + \sum_{v\in son(u)}f_v \cdot (1-\dfrac{p}{q}) \cdot \dfrac{1}{d_v} \]

（\(d_v\) 为 \(v\) 的度）
这样，我们只需要高斯消元，解出线性方程组即可。

代码

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 305, M = 90005;
const double eps = 1e-8;

int h[N], e[M], ne[M], idx;
long long inn[N];
int n, m, p, q;
double f[N], coeff[N][N];

void add(int a, int b){
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

void gauss(){
    for (int i = 1; i <= n; i ++ ){
        int row = 0;
        double maxn = -1e9;
        for (int j = i; j <= n; j ++ ) 
            if (abs(coeff[i][j]) > maxn) row = j, maxn = abs(coeff[i][j]);
        swap(coeff[i], coeff[row]);
        double Main = coeff[i][i];
        if (abs(Main) <= eps) {
            puts("NO SOLUTION");
            return;
        }
        for (int j = i; j <= n + 1; j ++ ) coeff[i][j] /= Main;
        for (int j = 1; j <= n; j ++ ){
            if (j == i) continue;
            double div = coeff[j][i];
            for (int k = i; k <= n + 1; k ++ ) coeff[j][k] -= div * coeff[i][k];
        }

    }
}

int main(){
    memset(h, -1, sizeof h);
    scanf("%d%d%d%d", &n, &m, &p, &q);
    while (m -- ){
        int u, v;
        scanf("%d%d", &u, &v);
        add(u, v), add(v, u);
        inn[u] ++ , inn[v] ++ ;
    }

    for (int u = 1; u <= n; u ++ ){
        if (u == 1) coeff[u][n + 1] = -1;
        coeff[u][u] = -1;
        for (int i = h[u]; ~i; i = ne[i]){
            int j = e[i];
            coeff[u][j] = 1.0 * (q - p) / (q * inn[j]);
        }   
    }

    gauss();

    for (int i = 1; i <= n; i ++ ) printf("%.9lf\n", coeff[i][n + 1] * p / q);

    return 0;
}