HDU 3853 LOOPS (概率DP)

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

令dp[i][j]表示从(i,j)到(R,C)花费的魔力值的期望。
那么，我们有：
dp[i][j] = 2 + ploop[i][j]*dp[i][j] + pleft[i][j]*dp[i][j+1] + pdown[i][j]*dp[i+1][j]
移项可得：

if (ploop[i][j] == 1) dp[i][j] = 0;
else dp[i][j] = (2 + pright[i][j] * dp[i][j + 1] + pdown[i][j] * dp[i + 1][j]) / (1 - ploop[i][j]);

故倒着递推。
最后输出f[0][0]。

完整代码：（用C++交会很快）

/*2703ms,31784KB*/

#include<cstdio>
const int N = 1005;

double ploop

, pright

, pdown

, dp

;

int main()
{
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				scanf("%lf%lf%lf", &ploop[i][j], &pright[i][j], &pdown[i][j]);
		for (int i = n - 1; i >= 0; i--)
			for (int j = m - 1; j >= 0; j--)
			{
				if (ploop[i][j] == 1) dp[i][j] = 0;
				else dp[i][j] = (2 + pright[i][j] * dp[i][j + 1] + pdown[i][j] * dp[i + 1][j]) / (1 - ploop[i][j]);
			}
		printf("%.3f\n", dp[0][0]);
	}
	return 0;
}