HDU 3853 LOOPS 概率dp（水

水水过~

#include <stdio.h>
#include <cstring>
#include <iostream>
#include <map>
#include <cmath>
template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
template <class T>
inline void pt(T x) {
    if (x <0) {
        putchar('-');
        x = -x;
    }
    if(x>9) pt(x/10);
    putchar(x%10+'0');
}
using namespace std;
typedef unsigned long long ull;
const double eps = 1e-10;
const int N = 1005;
int n, m;
double mp[N][N][3], dp[N][N];
double solve(){
	dp[n][m] = 0;
	for(int i = n; i ; i--)
		for(int j = m; j; j--)
		{
			if(i == n && j == m)continue;
			if(fabs(mp[i][j][0] - 1.0) < eps)
			{
				dp[i][j] = 0;
			}
			else
			{
				dp[i][j] = mp[i][j][1] * dp[i][j+1] + mp[i][j][2] * dp[i+1][j];
				dp[i][j] += 2.0;
				dp[i][j] /= (1 - mp[i][j][0]);
			}
		}
	return dp[1][1];
}
void input(){
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
			scanf("%lf %lf %lf", &mp[i][j][0], &mp[i][j][1], &mp[i][j][2]);
}
int main(){
    while(cin>>n>>m){
        input();
        printf("%.3f\n", solve());
    }
    return 0;
}