bzoj1778

高斯消元+矩阵的逆

来自popoqqq大神

求矩阵的逆：把I-T放在左边，P/Q*S放在右边，这样就形成了一个n*2n的矩阵，然后把左边高斯消元，右边就是求完逆的矩阵，其实就是ans，矩阵的逆跟乘法逆元是一样的，只不过是矩阵的逆元

然后输出a[i][n+1]，事实上矩阵只有n*(n+1)

构造转移概率矩阵是a[u][v]=1.0/d[v]*(1-p/q),就是v->u的概率乘上在v不爆炸的概率

#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-15;
const int N = 610;
int n, m;
double a[N][N];
double p, q, t;
vector<int> G[N];
void gauss_jordan()
{
    a[1][n + 1] = t;
    for(int i = 1; i <= n; ++i)
    {
        a[i][i] += 1.0;
        for(int j = 0; j < G[i].size(); ++j)
        {
            int u = G[i][j];
            a[i][u] -= (1.0 - t) / (double)(G[u].size());
        }
    }
    for(int now = 1; now <= n; ++now)
    {
        int x = now;
        for(int i = now; i <= n; ++i) if(fabs(a[i][now]) > fabs(a[x][now])) x = i;
        for(int i = 1; i <= n + 1; ++i) swap(a[now][i], a[x][i]);
        double t = a[now][now];
        for(int i = 1; i <= n + 1; ++i) a[now][i] /= t;
        for(int i = 1; i <= n; ++i) if(fabs(a[i][now]) > eps && now != i)
        {
            t = a[i][now];
            for(int j = 1; j <= n + 1; ++j) a[i][j] -= t * a[now][j];
        }
    }
    for(int i = 1; i <= n; ++i) printf("%.9f\n", a[i][n + 1]);
}
int main()
{
    scanf("%d%d%lf%lf", &n, &m, &p, &q);
    t = p / q;
    for(int i = 1; i <= m; ++i)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    gauss_jordan();
    return 0;
}