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
*/