Luogu

sol

首先解釋一波這道題無重邊無自環
設\(f_i\)表示\(i\)點上面的答案。
方程
\[f_u=\sum_{v,(u,v)\in E}(1-\frac PQ)\frac{f_v}{du_v}\]
\(f_1\)的那個方程加一個\(\frac PQ\)常數項

code

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int N = 305;
int gi()
{
    int x=0,w=1
 
;char ch=getchar();
    while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if (ch=='-') w=0,ch=getchar();
    while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
    return w?x:-x;
}
struct edge{int
 
 to,next;}a[N*N<<1];
int n,m,head[N],cnt;
double p,f[N][N],du[N],sol[N];
int main()
{
    n=gi();m=gi();
    p=(double)gi()/(double)gi();
    for (int i=1,u,v;i<=m;i++)
    {
        u=gi();v=gi();
        a[++cnt]=(edge){v,head[u]};head[u]=cnt;
        a[++cnt]=(edge){u,head[v]};head[v]=cnt;
        du[u]+=1.0
 
;du[v]+=1.0;
    }
    f[1][n+1]=p;
    for (int u=1;u<=n;u++)
    {
        f[u][u]=1;
        for (int e=head[u];e;e=a[e].next)
            f[u][a[e].to]-=(1-p)/du[a[e].to];
    }
    for (int i=1;i<=n;i++)
        for (int j=i+1;j<=n;j++)
            for (int k=n+1;k>=i;k--)
                f[j][k]-=f[i][k]*f[j][i]/f[i][i];
    for (int i=n;i;i--)
    {
        sol[i]=f[i][n+1];
        for (int j=n;j>i;j--)
            sol[i]-=f[i][j]*sol[j];
        sol[i]/=f[i][i];
    }
    for (int i=1;i<=n;i++) printf("%.9lf\n",sol[i]);
    return 0;
}

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