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

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

 int main(){
     freopen("bzoj2337.in","r",stdin);
     freopen("bzoj2337.out","w",stdout);
     scanf("%d%d",&n,&m);
     rep(i,,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=<<; i; i>>=) ans+=cal(i)*i;
     printf("%.3lf\n",ans);
     return ;
 }