bzoj 3143 [Hnoi2013]游走 期望dp+高斯消元

题面

题目传送门

解法

设\(f_i\)表示经过点\(i\)的期望次数

然后就可以通过边的关系搞出一个方程组，高斯消元一下

对于求一条边\((x,y)\)的概率，即为\(\frac{f_x}{out_x}+\frac{f_y}{out_y}\)

然后按照概率大小排序，小的用尽量大的编号

时间复杂度：\(O(n^3)\)

代码

#include <bits/stdc++.h>
#define double long double
#define eps 1e-9
#define N 510
using namespace std;
int s[N];
vector <int> e[N];
double b[N * N], a[N][N];
void gauss(int n) {
    for (int i = 1; i <= n; i++) {
        if (fabs(a[i][i]) < eps)
            for (int j = i + 1; j <= n; j++)
                if (fabs(a[j][i]) > eps)
                    for (int k = 1; k <= n + 1; k++) swap(a[i][k], a[j][k]);
        for (int j = i + 1; j <= n + 1; j++) a[i][j] /= a[i][i];
        for (int j = 1; j <= n; j++) {
            if (i == j) continue;
            for (int k = i + 1; k <= n + 1; k++)
                a[j][k] -= a[j][i] * a[i][k];
        }
    }
}
int main() {
    ios::sync_with_stdio(false);
    int n, m; cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int x, y; cin >> x >> y;
        s[x]++, s[y]++;
        e[x].push_back(y), e[y].push_back(x);
    }
    for (int i = 1; i < n; i++) {
        a[i][i] = 1;
        for (int j = 0; j < e[i].size(); j++) {
            int k = e[i][j];
            if (k != n) a[i][k] -= 1.0 / s[k];
        }
    }
    a[1][n] = 1;
    gauss(n - 1); int tot = 0;
    for (int i = 1; i < n; i++)
        for (int j = 0; j < e[i].size(); j++) {
            int k = e[i][j]; if (i > k) continue;
            b[++tot] = a[i][n] / s[i] + a[k][n] / s[k];
        }
    sort(b + 1, b + m + 1); double ans = 0;
    for (int i = 1; i <= m; i++) ans += b[i] * (m - i + 1);
    cout << fixed << setprecision(3) << ans << "\n";
    return 0;
}