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=0,f=1;char chr=getchar();
    while(!isdigit(chr)){if(chr=='-') f=-1;chr=getchar();}
    while(isdigit(chr)){ans=(ans<<3)+(ans<<1)+chr-48;chr=getchar();}
    return ans*f;
}void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}const int M = 305;
int n,m,p,q,du[M],e[M][M];
double a[M][M],ans[M],b[M];
const double eps=1e-13;
inline void Gauss(){
    for(int i=1;i<=n;i++){
        int maxn=i;
        for(int j=i+1;j<=n;j++) if(fabs(a[j][i]-a[maxn][i])<eps) maxn=j;
        for(int j=1;j<=n+1;j++) swap(a[maxn][j],a[i][j]);
        for(int j=n+1;j>=i;j--)
            for(int k=i+1;k<=n;k++)
                a[k][j]-=a[k][i]/a[i][i]*a[i][j];
    } 
    for(int i=n;i>=1;i--){ 
        for(int j=i+1;j<=n;j++)
            a[i][n+1]-=a[j][n+1]*a[i][j];
        a[i][n+1]/=a[i][i];
    }
    for(int i=1;i<=n;i++)    ans[i]=a[i][n+1];
}
int main(){
    n=read(),m=read(),p=read(),q=read();
    const double ps=p*1.0/q;
    for(int i=1;i<=m;i++){
        int x=read(),y=read();
        e[x][y]=e[y][x]=1;
        du[x]++,du[y]++;
    }
    for(int i=1;i<=n;i++){
        a[i][i]=1.0;
        for(int j=1;j<=n;j++)
            if(e[i][j])
                a[i][j]-=(1.0-ps)/du[j];
    }
    a[1][n+1]=1.0;
    Gauss();
    for(int i=1;i<=n;i++)printf("%.9lf\n",ans[i]*ps);
    return 0;
}