zoj 3822 Domination(dp)，zoj3822

zoj 3822 Domination(dp)，zoj3822

題目連結：zoj 3822 Domination

題目大意：給定一個N∗M的棋盤，每次任選一個位置放置一枚棋子，直到每行每列上都至少有一枚棋子，問放置棋子個數的期望。

解題思路：大白書上機率那一張有一道類似的題目，但是因為時間比較久了，還是稍微想了一下。
dp[i][j][k]表示i行j列上均有至少一枚棋子，並且消耗k步的機率（k≤i∗j）,因為放置在i+1~n上等價與放在i+1行上，同理列也是如此。所以有轉移方程：

• dp[i][j][k+1]+=dp[i][j][k]∗(n−k)(S−k)
• dp[i+1][j][k+1]+=dp[i][j][k]∗(N−i)∗j(S−k)
• dp[i][j+1][k+1]+=dp[i][j][k]∗(M−j)∗i(S−k)
• dp[i+1][j+1][k+1]+=dp[i][j][k]∗(N−i)∗(M−j)(S−k)

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 55;const int maxm = 2505;int N, M;double dp[maxn][maxn][maxm];double solve () {    int S = N * M;    memset(dp, 0, sizeof(dp));    dp[1][1][1] = 1;    for (int i = 1; i <= N; i++) {        for (int j = 1; j <= M; j++) {            int n = i * j;            for (int k = max(i, j); k <= n; k++) {                dp[i][j][k+1] += dp[i][j][k] * (n - k) / (S - k);                dp[i+1][j][k+1] += dp[i][j][k] * (N - i) * j / (S - k);                dp[i][j+1][k+1] += dp[i][j][k] * (M - j) * i / (S - k);                dp[i+1][j+1][k+1] += dp[i][j][k] * (N - i) * (M - j) / (S - k);            }        }    }    /*    for (int i = 1; i <= N; i++) {        for (int j = 1; j <= M; j++) {            printf("%d %d:", i, j);            for (int k = max(i, j); k <= i * j; k++)                printf("%.3lf ", dp[i][j][k]);            printf("\n");        }    }    */    double ans = 0;    for (int i = max(N, M); i <= S; i++)        ans += (dp[N][M][i] - dp[N][M][i-1]) * i;    return ans;}int main () {    int cas;    scanf("%d", &cas);    while (scanf("%d%d", &N, &M) == 2) {        printf("%.8lf\n", solve());    }    return 0;}