bzoj 2337 [HNOI2011]XOR和路径 高斯消元+期望dp

题面

题目传送门

解法

既然有异或，那么我们把每一位单独考虑一下

先枚举是二进制的第几位，然后设\(f_i\)表示点\(i\)这一位为1的概率是多少

显然，可以列出一个方程

注意，自环的出度不能被计算2遍

高斯消元解这个方程即可

最后答案为\(\sum2^i×f_{1,i}\)

时间复杂度：\(O(30n^3)\)

代码

#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;
}