LOOPS HDU-3853 期望DP

LOOPS

solution

$$dp[i][j]=dp[i][j]\ast p[i][j][1]+dp[i][j-1]\ast p[i][j][2]+dp[i-1][j]\ast p[i][j][3]$$

code

/*Siberian Squirrel*/
/*Cute KiloFish*/
#include<bits/stdc++.h>

#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ACM_LOCAL

using namespace std;
typedef long long ll;

const double PI = acos(-1);
const double eps = 1e-4;
/*const int MOD = 998244353, r = 119, k = 23, g = 3;
const int MOD = 1004535809, r = 479, k = 21, g = 3;*/
const int MOD = 1e9 + 7;
const int M = 1e3 + 10;
const int N = 3e5 + 10;

//inline int rnd(){static int seed=2333;return seed=(((seed*666666ll+20050818)%998244353)^1000000007)%1004535809;}

int r, c;
double dp[M][M];
double p[M][M][4];

void solve() {
    
    
    for(int i = 1; i <= r; ++ i)
        for(int j = 1; j <= c; ++ j)
            for(int k = 1; k <= 3; ++ k)
                scanf("%lf", &p[i][j][k]);
    for(int i = r; i >= 1; -- i) {
    
    
        for(int j = c; j >= 1; -- j) {
    
    
            if(!(i == r && j == c)) {
    
    
                if (fabs(p[i][j][1] - 1.0) > eps) {
    
    
                    dp[i][j] = (dp[i][j + 1] * p[i][j][2] + dp[i + 1][j] * p[i][j][3] + 2) / (1 - p[i][j][1]);
                }
            }
        }
    }
    printf("%.3f\n", dp[1][1]);
}

int main() {
    
    
    IO;
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif
    int o = 1, cases = 0;
//    cin >> o;
    while(o --) {
    
    
        while(~scanf("%d%d", &r, &c)) {
    
    
            dp[r][c] = 0;
            solve();
        }
    }
    return 0;
}