纪念首道期望题（虽说绿豆蛙的归宿才是，但是我打的深搜总觉得不正规）。
我们求出每条边的期望经过次数，然后排序，经过多的序号小，经过少的序号大，这样就可以保证最后的值最小。
对于每一条边的期望经过次数，其实是从它起点和终点来的。设f[]为每个点经过的期望,out[]为每个点的出度
设一条边，起点为u，终点为v。那么它的期望经过次数为f[u]/out[u]+f[v]/out[v]
这样问题就转化为求每个点的期望经过次数了
对于起点,一开始经过一次,也可能从其他点走过来.
f[1]=1+sigma(f[j]/out(j),j和1有边)
f[i]=sigma(f[j]/out(j),j和i有边) //i>=2
这是n个变量n个方程的方程组,高斯消元解方程组,O(n^3)
#include<iostream>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;#define pos(i,a,b) for(int i=(a);i<=(b);i++)#define pos2(i,a,b) for(int i=(a);i>=(b);i--)#define N 510int n,m;int read(){ int su=0; char ch=getchar(); while(ch<'0'||ch>'9') ch=getchar(); while(ch<='9'&&ch>='0'){su=su*10+ch-'0';ch=getchar(); } return su;}double out[N];double a[N][N],b[N],x[N];int lian[N][N];void swap(double &xx,double &yy){ double temp; temp=xx; xx=yy; yy=temp;}void gauss(){ double t; int im,num=1; for(int k=1;k<n;k++,num++) { im=k; pos(i,k+1,n) if(fabs(a[i][k])>fabs(a[im][k])) im=i; if(im!=k){ pos(i,k,n) swap(a[num][i],a[im][i]); swap(b[num],b[im]); } if(!a[num][k]){ num--; continue; } pos(i,num+1,n){ if(!a[num][k]) continue; t=a[i][k]/a[num][k]; pos(j,k,n+1) a[i][j]-=t*a[k][j]; b[i]-=t*b[k]; } } pos2(i,n,1){ pos2(j,n,i+1) b[i]-=a[i][j]*x[j]; x[i]=b[i]/a[i][i]; }}struct qian{int from,to;double e;}cun[N*N];bool aaa(const qian &a,const qian &b){return a.e<b.e;}double ans;int messi(){freopen("walk.in","r",stdin);freopen("walk.out","w",stdout);n=read();m=read();pos(i,1,m){int x,y;x=read();y=read();lian[x][y]=lian[y][x]=1;out[x]+=1.0;out[y]+=1.0;cun[i].from=x;cun[i].to=y;}out[n]=0.0;a[1][1]=-1.0;b[1]=-1.0;pos(i,2,n){if(lian[i][1]==1&&out[i]){a[1][i]=1.0/out[i];}}pos(i,2,n){a[i][i]=-1.0;pos(j,1,n){if(j!=i&&lian[j][i]==1&&out[j]){a[i][j]=1.0/out[j];}}}gauss();pos(i,1,m){if(out[cun[i].from]!=0&&out[cun[i].to]!=0)cun[i].e=x[cun[i].from]/out[cun[i].from]+x[cun[i].to]/out[cun[i].to];else{if(out[cun[i].from]==0) cun[i].e=x[cun[i].to]/out[cun[i].to];else cun[i].e=x[cun[i].from]/out[cun[i].from];}}sort(cun+1,cun+m+1,aaa);pos(i,1,m){ans+=cun[i].e*(double)(m-i+1);}printf("%0.3lf",ans);return 0;}int hallmeow=messi();int main(){;}
