題面

題目傳送門

解法

既然有異或，那麽我們把每一位單獨考慮一下

先枚舉是二進制的第幾位，然後設\(f_i\)表示點\(i\)這一位為1的概率是多少

顯然，可以列出一個方程

註意，自環的出度不能被計算2遍

高斯消元解這個方程即可

最後答案為\(\sum2^i×f_{1,i}\)

時間復雜度：\(O(30n^3)\)

代碼

#include <bits/stdc++.h>
#define double long double
#define eps 1e-9
#define N 110
using namespace std;
struct Edge {
    int next, num, v;
} e[N * N * 16];
int cnt, s[N];
double a[N][N];
void add(int x, int y, int v) {
    e[++cnt] = (Edge) {e[x].next, y, v};
    e[x].next = cnt;
}
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; cnt = n;
    for (int i = 1; i <= m; i++) {
        int x, y, v;
        cin >> x >> y >> v;
        if (x == y) s[x]++, add(x, y, v);
            else s[x]++, s[y]++, add(x, y, v), add(y, x, v);
    }
    double ans = 0;
    for (int l = 0; l <= 30; l++) {
        memset(a, 0, sizeof(a));
        for (int i = 1; i < n; i++) {
            a[i][i] = 1;
            for (int p = e[i].next; p; p = e[p].next) {
                int k = e[p].num, v = e[p].v;
                if (((v >> l) & 1) == 1) {
                    a[i][k] += 1.0 / s[i];
                    if (k != n) a[i][n] += 1.0 / s[i];
                } else if (k != n) a[i][k] -= 1.0 / s[i];
            }
        }
        gauss(n - 1); ans += (1 << l) * a[1][n];
    }
    cout << fixed << setprecision(3) << ans << "\n";
    return 0;
}
 

bzoj 2337 [HNOI2011]XOR和路徑 高斯消元+期望dp