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HDU_3853_概率dp

2024-08-14 02:57:00   219人阅读

http://acm.hdu.edu.cn/showproblem.php?pid=3853

 

又因为总期望为子期望的加权和，加权因子为子期望的转移概率，所以得到：
dp[ i ][ j ]= p1 * ( dp[ i ][ j ] + 2 ) + p2 * ( dp[ i ][ j + 1 ] + 2 ) + p3 * ( dp[ i + 1 ][ j ] + 2) 。

 

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int c,r;double ans[1005][1005],stay[1005][1005],rightt[1005][1005],down[1005][1005];int main(){    while(~scanf("%d%d",&r,&c))    {        memset(ans,0,sizeof(ans));        for(int i = 1;i <= r;i++)        {            for(int j = 1;j <= c;j++)   scanf("%lf%lf%lf",&stay[i][j],&rightt[i][j],&down[i][j]);        }        for(int i = r;i >= 1;i--)        {            for(int j = c;j >= 1;j--)            {                if(-1e-6 <= stay[i][j]-1 && stay[i][j]-1 <= 1e-6)   continue;                ans[i][j] = (2+rightt[i][j]*ans[i][j+1]+down[i][j]*ans[i+1][j])/(1-stay[i][j]);            }        }        printf("%.3f\n",ans[1][1]);    }    return 0;}

 

HDU_3853_概率dp


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