HDU-3853 LOOPS 概率DP

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

　　简单概率DP，转移方程：f[i][j]=f[i][j]*p1+f[i][j+1]*p2+f[i+1][j]*p3+2  —>  f[i][j]=(f[i][j+1]*p2+f[i+1][j]*p3+2)/(1-p1).

//STATUS:C++_AC_2828MS_32180KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=1010;
const int INF=0x3f3f3f3f;
const int MOD= 1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e30;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
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));}
//End

double f[N][N],p[N][N][3];
int n,m;

double dfs(int x,int y)
{
    if(f[x][y]>=0)return f[x][y];
    if(x==n-1 && y==m-1)return f[x][y]=0;
    f[x][y]=2/(1-p[x][y][0]);
    if(p[x][y][1]>0)
        f[x][y]+=dfs(x,y+1)*p[x][y][1]/(1-p[x][y][0]);
    if(p[x][y][2]>0)
        f[x][y]+=dfs(x+1,y)*p[x][y][2]/(1-p[x][y][0]);
    return f[x][y];
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        mem(p,0);
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                scanf("%lf%lf%lf",&p[i][j][0],&p[i][j][1],&p[i][j][2]);
            }
        }

        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
                f[i][j]=-1;
        dfs(0,0);

        printf("%.3lf
",f[0][0]);
    }
    return 0;
}