P2973 [USACO10HOL]赶小猪

跟那个某省省选题(具体忘了)游走差不多...

把边搞到点上然后按套路Gauss即可

貌似有人说卡精度，$eps≤1e-13$，然而我$1e-12$也可以过...

代码：

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #define writeln(x)  write(x),puts("")
 #define writep(x)   write(x),putchar(' ')
 using namespace std;
 inline int read(){
     int ans=,f=;char chr=getchar();
     while(!isdigit(chr)){if(chr=='-') f=-;chr=getchar();}
     while(isdigit(chr)){ans=(ans<<)+(ans<<)+chr-;chr=getchar();}
     return ans*f;
 }void write(int x){
     if(x<) putchar('-'),x=-x;
     if(x>) write(x/);
     putchar(x%+'');
 }const int M = ;
 int n,m,p,q,du[M],e[M][M];
 double a[M][M],ans[M],b[M];
 const double eps=1e-;
 inline void Gauss(){
     for(int i=;i<=n;i++){
         int maxn=i;
         for(int j=i+;j<=n;j++) if(fabs(a[j][i]-a[maxn][i])<eps) maxn=j;
         for(int j=;j<=n+;j++) swap(a[maxn][j],a[i][j]);
         for(int j=n+;j>=i;j--)
             for(int k=i+;k<=n;k++)
                 a[k][j]-=a[k][i]/a[i][i]*a[i][j];
     }
     for(int i=n;i>=;i--){
         for(int j=i+;j<=n;j++)
             a[i][n+]-=a[j][n+]*a[i][j];
         a[i][n+]/=a[i][i];
     }
     for(int i=;i<=n;i++)    ans[i]=a[i][n+];
 }
 int main(){
     n=read(),m=read(),p=read(),q=read();
     const double ps=p*1.0/q;
     for(int i=;i<=m;i++){
         int x=read(),y=read();
         e[x][y]=e[y][x]=;
         du[x]++,du[y]++;
     }
     for(int i=;i<=n;i++){
         a[i][i]=1.0;
         for(int j=;j<=n;j++)
             if(e[i][j])
                 a[i][j]-=(1.0-ps)/du[j];
     }
     a[][n+]=1.0;
     Gauss();
     for(int i=;i<=n;i++)printf("%.9lf\n",ans[i]*ps);
     return ;
 }