[BZOJ2337][HNOI2011]XOR和路径(概率+高斯消元)

直接不容易算，考虑拆成位处理。

设f[i]表示i到n的期望路径异或和（仅考虑某一位），则$f[y]=\sum\limits_{exist\ x1\to y=0}\frac{f[x1]}{d[x1]}+\sum\limits_{exist\ x2\to y=1}\frac{1-f[x2]}{d[x2]}$。

对于重边，直接在系数上+1即可。对于自环，只计算一次度数即可。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,l,r) for (int i=l; i<=r; i++)
typedef long long ll;
using namespace std;

const int N=110,M=10010;
int n,m,d[N];
double ans,a[N][N];
struct E{ int x,y,w; }e[M];
double Abs(double x){ return (x<0) ? -x : x; }

double cal(int S){
    memset(a,0,sizeof(a));
    rep(i,1,m){
        if (e[i].w&S){
            a[e[i].x][e[i].y]+=1; a[e[i].x][n+1]+=1;
            if (e[i].x!=e[i].y) a[e[i].y][e[i].x]+=1,a[e[i].y][n+1]+=1;
        }else{
            a[e[i].x][e[i].y]-=1;
            if (e[i].x!=e[i].y) a[e[i].y][e[i].x]-=1;
        }
    }
    rep(i,1,n+1) a[n][i]=0;
    rep(i,1,n) a[i][i]+=d[i];
    rep(i,1,n){
        int k=i;
        rep(j,i+1,n) if (Abs(a[j][i])>Abs(a[k][i])) k=j;
        rep(j,i,n+1) swap(a[k][j],a[i][j]);
        rep(j,i+1,n){
            double t=a[j][i]/a[i][i];
            rep(k,i,n+1) a[j][k]-=a[i][k]*t;
        }
    }
    for (int i=n; i; i--){
        rep(j,i+1,n) a[i][n+1]-=a[i][j]*a[j][n+1];
        a[i][n+1]/=a[i][i];
    }
    return a[1][n+1];
}

int main(){
    freopen("bzoj2337.in","r",stdin);
    freopen("bzoj2337.out","w",stdout);
    scanf("%d%d",&n,&m);
    rep(i,1,m){
        scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w); d[e[i].x]++;
        if (e[i].y!=e[i].x) d[e[i].y]++;
    }
    for (int i=1<<30; i; i>>=1) ans+=cal(i)*i;
    printf("%.3lf\n",ans);
    return 0;
}