HDU3853:LOOPS

题意：迷宫是一个R*C的布局，每个格子中给出停留在原地，往右走一个，往下走一格的概率，起点在(1,1)，终点在(R,C)，每走一格消耗两点能量，求出最后所需要的能量期望

 

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
typedef long long ll;
const int inf=0x3f3f3f3f;
#define For(i,a,b) for(int i=a;i<=b;i++)
#define FFor(i,a,b) for(int i=a;i>=b;i--)
template <class T> inline T min(T a,T b,T c)
{
    return min(min(a,b),c);
}
template <class T> inline T max(T a,T b,T c)
{
    return max(max(a,b),c);
}
template <class T> inline T min(T a,T b,T c,T d)
{
    return min(min(a,b),min(c,d));
}
template <class T> inline T max(T a,T b,T c,T d)
{
    return max(max(a,b),max(c,d));
}
using namespace std;
const int N=1e3+5;
const double eps=1e-9;
//***********name**************
double f[N][N];
int R,C;
double p0[N][N],p1[N][N],p2[N][N];

//***********function**********

//******************************
int main()
{
//    freopen("test.txt","r",stdin);
while(~scanf("%d%d",&R,&C)){
memset(f,0,sizeof(f));
    For(i,1,R)
    For(j,1,C)
    scanf("%lf%lf%lf",&p0[i][j],&p1[i][j+1],&p2[i+1][j]);
    f[R][C]=0;
    FFor(i,R,1)
    FFor(j,C,1)
    {
        if(i==R&&j==C)continue;
        if(fabs(p0[i][j]-1)<=eps)continue;
        f[i][j]=(f[i][j+1]*p1[i][j+1]+f[i+1][j]*p2[i+1][j]+2)/(1-p0[i][j]);
    }
    printf("%.3f\n",f[1][1]);
}

    return 0;
}