[Luogu2973][USACO10HOL]赶小猪

Luogu

sol

首先解释一波这道题无重边无自环
设\(f_i\)表示\(i\)点上面的答案。
方程

\[f_u=\sum_{v,(u,v)\in E}(1-\frac PQ)\frac{f_v}{du_v} \]

\(f_1\)的那个方程加一个\(\frac PQ\)常数项

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;
}