HDU5245(概率DP)

来源：互联网发布：侠客风云传优化补丁编辑：程序博客网时间：2024/05/29 02:18

题意是在一个m×n的矩阵李随机选择两个点作为一个子矩阵斜边端点并且给这个子矩阵涂颜色。问k次涂色后矩阵中涂色的方块的个数的期望。

方块个数的期望就是k步后每个方块已经被涂色的概率的和。

预先计算出每个方块一次中被涂色的概率。

坑点就是小心爆int，我一气之下都改成了long long。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>using namespace std;#define maxn 511double f[maxn][maxn], dp[maxn][maxn][22];long long n, m, k;int main () {    int t, kase = 0;    cin >> t;    while (t--) {        cin >> n >> m >> k;        for (long long i = 1; i <= n; i++) {            for (long long j = 1; j <= m; j++) {                long long ans1 = (i-1)*m*(i-1)*m+(n-i)*m*(n-i)*m+(j-1)*n*(j-1)*n+(m-j)*n*(m-j)*n;                ans1 = ans1 - (i-1)*(j-1)*(i-1)*(j-1) - (i-1)*(m-j)*(i-1)*(m-j) - (n-i)*(j-1)*(n-i)*(j-1) - (n-i)*(m-j)*(n-i)*(m-j);                long long ans2 = m*n*m*n;                f[i][j] = (ans2-ans1)*1.0/ans2;            }        }        for (long long i = 1; i <= n; i++) {            for (long long j = 1; j <= m; j++) {                dp[i][j][0] = 0.0;                for (long long x = 1; x <= k; x++) {                    dp[i][j][x] = dp[i][j][x-1] + (1.0-dp[i][j][x-1])*f[i][j];                }            }        }        double ans = 0.0;        for (long long i = 1; i <= n; i++) {            for (long long j = 1; j <= m; j++) {                ans += dp[i][j][k];            }        }        printf ("Case #%d: %lld\n", ++kase, (long long) (ans + 0.5));    }    return 0;}

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