Luogu P2973 [USACO10HOL]赶小猪Driving Out the Piggi 后效性DP

Luogu P2973 [USACO10HOL]赶小猪Driving Out the Piggi 后效性DP

有后效性的DP：$f[u]$表示到$u$的期望次数，$f[u]=Sigma_{(u,v)} (1-frac{p}{q})*f[v]*deg[v]$，最后答案就是$f[u]*p/q$

刚开始$f[1]=1$,，因为炸弹初始在$1$号节点。所以增广矩阵中$a[1][n+1]=1$。

系数矩阵$a[i][i]$赋值为1，其他点的系数写成负数,相当于是所有的加起来$=0$.

#include<cstdio>
#include<iostream>
#include<cmath>
#define R register int
#define db long double
using namespace std;
const db eps=1E-13;
const int N=310;
db a[N][N],P;
int mp[N][N],r[N],n,m,p,q;
inline int g() {R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
}
inline bool ck0(const db& x) {return x<eps&&x>-eps;}
inline void Gauss(int n) {for(R i=1;i<=n;++i) { R p=i;for(R j=i+1;j<=n;++j) if(ck0(a[p][i])||fabs(a[p][i])<fabs(a[j][i])) p=j;if(p!=i) swap(a[p],a[i]);for(R j=i+1;j<=n;++j) if(!ck0(a[j][i])) {register db t=a[j][i]/a[i][i];for(R k=1;k<=n+1;++k) a[j][k]-=t*a[i][k];}} for(R i=n;i>=1;--i) {for(R j=n;j>i;--j) a[i][n+1]-=a[i][j]*a[j][n+1];a[i][n+1]/=a[i][i];}
}
signed main() {n=g(),m=g(),p=g(),q=g(); P=(db)p/q;for(R i=1,u,v;i<=m;++i) u=g(),v=g(),mp[u][v]=mp[v][u]=1,++r[u],++r[v];for(R i=1;i<=n;++i) a[i][i]=1;for(R i=1;i<=n;++i) for(R j=1;j<=n;++j) if(mp[i][j]) a[i][j]-=(1.0-P)/r[j];a[1][n+1]=1; Gauss(n); for(R i=1;i<=n;++i) printf("%.9Lfn",a[i][n+1]*P);
} 

2019.05.24

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