BZOJ2337: [HNOI2011]XOR和路径(期望 高斯消元)

题意

题目链接

sol

期望的线性性对xor运算是不成立的，但是我们可以每位分开算

设\(f[i]\)表示从\(i\)到\(n\)边权为1的概率，统计答案的时候乘一下权值

转移方程为

\[f[i] = (w = 1) \frac{1 - f[to]}{deg[i]} +(w = 0) \frac{f[to]}{deg[i]} \]

高斯消元解一下

注意:f[n] = 0,有重边！

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1001;
inline int read() {
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, deg[maxn];
vector<int> a[maxn][maxn];
double f[maxn][maxn];
void gauss() {
    for(int i = 1; i <= n - 1; i++) {
        int mx = i;
        for(int j = i + 1; j <= n; j++) if(f[j][i] > f[mx][i]) mx = j;
        if(mx != i) swap(f[i], f[mx]);
        for(int j = 1; j <= n; j++) {
            if(i == j) continue;
            double p = f[j][i] / f[i][i];
            for(int k = i; k <= n + 1; k++) f[j][k] -= f[i][k] * p;
        }
    }
    for(int i = 1; i <= n; i++) f[i][n + 1] /= f[i][i];
}
int main() {
    //freopen("2.in", "r", stdin);
    n = read(); m = read();
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read(), z = read();
        a[x][y].push_back(z); 
        deg[x]++;
        if(x != y) deg[y]++, a[y][x].push_back(z);
    }
    double ans = 0;
    for(int b = 0; b <= 31; b++) {
        memset(f, 0, sizeof(f));
        for(int i = 1; i <= n - 1; i++) {
            f[i][i] = deg[i];
            for(int j = 1; j <= n; j++) {
                for(int k = 0; k < a[i][j].size(); k++) {
                    int w = a[i][j][k];
                    if(w & (1 << b)) {//
                        f[i][n + 1]++;
                        if(j != n) f[i][j]++;
                    } else {
                        if(j != n) f[i][j]--;
                    }
                }
            }
        }
        gauss();
        ans += (1 << b) * f[1][n + 1];
    }
    printf("%.3lf", ans);
    return 0;
}
/*
3 3 
1 2 4 
1 3 5 
2 3 6
*/