bzoj4008

期望DP和概率DP

Fi,j×Fi,j×

(1−pi+1)j−>Fi+1,j(1−pi+1)j−>Fi+1,j

1−(1−pi+1)j−>Fi+1,j−1

#include<cstdio>
#include<cctype>
#include<cstring>
using namespace std;
int n,r,d[222];
double p[222],f[222][134],pw[222][134];
//f[i][j]表示当前到了第i个(发动了技能)，还剩j轮未发动技能时的概率 
inline void read(int &x){
    char ch=getchar();x=0;
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
}

int main(){
    int T;read(T);
    while(T--){
        read(n);read(r);
        memset(f,0,sizeof(f));
        for(int i=1;i<=n;i++){
            scanf("%lf",&p[i]);
            read(d[i]);
            pw[i][0]=1.0;
            for(int j=1;j<=r;j++)pw[i][j]=pw[i][j-1]*(1-p[i]);//pw[i][j]计算(1-p[i]) ^j,即接下来j轮均不被发动的概率 
        }
        double ans=0;
        f[0][r]=1;
        for(int i=0;i<n;i++)
           for(int j=0;j<=r;j++){
               f[i+1][j]+=f[i][j]*pw[i+1][j];
               if(j-1>=0){
                   f[i+1][j-1]+=f[i][j]*(1-pw[i+1][j]);//(1-pw[i+1][j]表示当前第j+1在接下来j轮中发动技能的概率 
                   ans+=f[i][j]*(1-pw[i+1][j])*d[i+1];//因为只需加上当前一轮的，因此不是f[i+1][j-1]*... 
               }
           }
        printf("%.10lf\n",ans);
    }
}