[cf113d]Museum

传送门

Solution

设一个状态为 \((x,y)\) 表示两人在的位置，求出每个状态期望出现的次数

设一个状态为 \(u\) , \(x_u^0=[u==(a,b)]\)

所以一个状态出现的次数期望为 \(E_u=x_u^0+\sum_vG(v,u)\times E_v\)

其中\(G\)是状态转移的概率矩阵

高消即可，复杂度 \(O(n^6)\)

Code 

#include<bits/stdc++.h>
#define ll long long
#define db double
#define dbg1(x) cerr<<#x<<"="<<(x)<<" "
#define dbg2(x) cerr<<#x<<"="<<(x)<<"\n"
#define dbg3(x) cerr<<#x<<"\n"
using namespace std;
#define reg register
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
const int MN=23;bool A[MN][MN];
int I[MN][MN],N,M,a,b,deg[MN];
db B[MN*MN][MN*MN],F[MN*MN],P[MN];
void Gauss(int n)
{
    reg int i,j,k;
    for(i=1;i<=n;++i)
    {
        int p=i;
        for(j=i+1;j<=n;++j)if(fabs(B[j][i])>fabs(B[p][i]))p=j;
        if(i!=p)swap(B[i],B[p]);
        for(j=i+1;j<=n;++j)
        {
            db d=B[j][i]/B[i][i];
            for(k=i;k<=n+1;++k) B[j][k]-=d*B[i][k];
        }
	}
    for(i=n;i;--i)
    {
        for(j=i+1;j<=n;++j)B[i][n+1]-=F[j]*B[i][j]; 
        F[i]=B[i][n+1]/B[i][i];
    }
}
db _(int x,int y){if(!A[x][y])return 0;if(x==y)return P[x];return (1.-P[x])/deg[x];}
int main()
{
	reg int i,j,x,y;
	N=read(),M=read();a=read(),b=read();
	for(i=1;i<=M;++i) x=read(),y=read(),++deg[x],++deg[y],A[x][y]=A[y][x]=1;
	for(i=1;i<=N;++i) A[i][i]=1,scanf("%lf",&P[i]);
	for(i=1;i<=N;++i)for(j=1;j<=N;++j)I[i][j]=(i-1)*N+j;
	B[I[a][b]][N*N+1]=-1.;
	for(i=1;i<=N;++i)for(j=1;j<=N;++j)
	{
		B[I[i][j]][I[i][j]]+=-1.;
		if(i!=j)for(x=1;x<=N;++x)for(y=1;y<=N;++y) B[I[x][y]][I[i][j]]+=_(i,x)*_(j,y);
	}
	Gauss(N*N);
	for(i=1;i<=N;++i) printf("%.10lf ",F[I[i][i]]);
	return 0;
}

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