BZOJ 3143: [Hnoi2013]游走

来源：互联网发布：电子手抄报软件编辑：程序博客网时间：2024/04/29 23:20

HNOI2013 get!

高斯消元求出每个点经过的次数的期望，然后算出每条边的期望

然后套排序原理，乱序和最小

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define double long doubleconst int N=500+5;const double eps=1e-9;double g[N][N];void Gauss(int n){for(int i=1;i<=n;i++){int j;for(j=i;j<=n;j++)if(fabs(g[j][i])>eps)break;for(int k=i;k<=n;k++)swap(g[i][k],g[j][k]);for(j=i+1;j<=n;j++){double t=g[j][i]/g[i][i];for(int k=i;k<=n+1;k++)g[j][k]-=t*g[i][k];}}for(int i=n;i>=1;i--){for(int j=i+1;j<=n;j++)g[i][n+1]-=g[j][n+1]*g[i][j];g[i][n+1]/=g[i][i];}}bool e[N][N];double deg[N];double path[N*N];int main(){//freopen("a.in","r",stdin);int n,m;scanf("%d%d",&n,&m);while(m--){int u,v;scanf("%d%d",&u,&v);e[u][v]=e[v][u]=1;deg[u]+=1.0;deg[v]+=1.0;}for(int i=1;i<=n;i++){g[i][i]=1.0;for(int j=1;j<=n;j++)if(e[j][i])g[i][j]-=1.0/deg[j];g[i][n]=0.0;}g[n][n]=0.0;g[1][n]=1.0;Gauss(n-1);double ans=0.0;int tot=0;for(int i=1;i<=n;i++)for(int j=i+1;j<=n;j++)if(e[i][j])path[++tot]=g[i][n]/deg[i]+g[j][n]/deg[j];sort(path+1,path+1+tot);for(int i=1;i<=tot;i++)ans+=path[i]*(tot-i+1);printf("%.3Lf\n",ans);return 0;}

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