嘟嘟嘟

一看到异或，就想到按位处理．
当处理到第\(i\)位的时候，\(f[u]\)表示节点\(u\)到\(n\)的路径，这一位为\(1\)的期望，那么为\(0\)就是\(1 - f[u]\)，于是有
\[f[u] = \frac{1}{d[u]} (\sum _ {v \in V, w = 0} f[v] + \sum _ {v \in V, w = 1} 1 - f[v])\]
因为是异或，所以如果边权这一位是0的话，应该加上\(f[v]\)；否则加上\(1 - f[v]\)。
然后整理一下
\[d[u] * f[u] - \sum _ {v \in V, w = 0} f[v] + \sum _ {v \in V, w = 1} f[v] = \sum _ {v \in V, w = 1} 1\]
于是就可以高斯消元了。
答案为\(\sum 2 ^ i * ans_i[1]\)。

需要注意的是,重边只应该加一次,对应的度数也应该只加\(1\)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 105;
const int maxe = 2e4 + 5;
inline ll read()
{ll ans = 0;char ch = getchar(), last = ' ';while(!isdigit(ch)) last = ch, ch = getchar();while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();if(last == '-') ans = -ans;return ans;
}
inline void write(ll x)
{if(x < 0) x = -x, putchar('-');if(x >= 10) write(x / 10);putchar(x % 10 + '0');
}int n, m, Max = 0;
int du[maxn];
struct Edge
{int nxt, to, w;
}e[maxe];
int head[maxn], ecnt = -1;
void addEdge(int x, int y, int w)
{e[++ecnt] = (Edge){head[x], y, w};head[x] = ecnt;
}db f[maxn][maxn], ans[maxn], Ans = 0;
void build(int x)
{Mem(f, 0);for(int i = 1; i < n; ++i)  //小于n{f[i][i] = du[i];for(int j = head[i]; j != -1; j = e[j].nxt)if((e[j].w >> x) & 1) ++f[i][e[j].to], ++f[i][n + 1];else --f[i][e[j].to];}
}
db Gauss()
{for(int i = 1; i <= n; ++i){int pos = i;for(int j = i + 1; j <= n; ++j)if(fabs(f[j][i]) > fabs(f[pos][i])) pos = j;if(pos != i) swap(f[i], f[pos]);db tp = f[i][i];if(fabs(tp) > eps) for(int j = i; j <= n + 1; ++j) f[i][j] /= tp;for(int j = i + 1; j <= n; ++j){db tp = f[j][i];for(int k = i; k <= n + 1; ++k) f[j][k] -= tp * f[i][k];}}for(int i = n; i; --i){ans[i] = f[i][n + 1];for(int j = i - 1; j; --j) f[j][n + 1] -= f[j][i] * f[i][n + 1];}return ans[1];
}int main()
{Mem(head, -1);n = read(); m = read();for(int i = 1; i <= m; ++i){int x = read(), y = read(), w = read();addEdge(x, y, w); ++du[x];if(x ^ y) addEdge(y, x, w), ++du[y];Max = max(Max, w);}for(int i = 0; (1 << i) <= Max; ++i)build(i), Ans += Gauss() * (1 << i);printf("%.3lf\n", Ans);return 0;
}