[黑书 - 动态规划] 例题2 - 棋盘分割

最新推荐文章于 2022-08-23 19:38:13 发布

ACMega

最新推荐文章于 2022-08-23 19:38:13 发布

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分类专栏：动态规划黑书 POJ

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本文详细解析了POJ1191-棋盘分割问题的动态规划算法实现过程，介绍了状态转移方程及边界条件，并给出了完整的C++代码实现。

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POJ1191 - 棋盘分割

状态转移方程：

Dp( x1, y1, x2, y2, k ) 是指矩形 (x1, y1, x2, y2) 切 k 次后各矩形总平方和的最小值

Dp( x1, y1, x2, y2, k ) = min { Dp( x1, y1, a, y2, k-1 ) + cal(a+1, y1, x2, y2 ), Dp( a+1, y1, x2, y2, k-1 ) + cal( x1, y1, a, y2 ) } 横切

min { Dp( x1, y1, x2, b, k-1 ) + cal(x1, b+1, x2, y2 ), Dp( x1, b+1, x2, y2, k-1 ) + cal( x1, y1, x2, b ) } 纵切

边界条件：

Dp( x1, y1, x2, y2, k ) = cal( x1, y1, x2, y2 ) if ( x1 = x2 | y1 = y2 | k = 1)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define INF (int)(1e9)
#define MAXN 8
using namespace std;
int mat[9][9];
int dp[15][9][9][9][9];
int cal(int x1, int y1, int x2, int y2) {
    int ret = (mat[x2][y2] - mat[x2][y1-1] - mat[x1-1][y2] + mat[x1-1][y1-1]);
    return ret*ret;
}
int solve(int x1, int y1, int x2, int y2, int k) {
    int a, b;
    if (dp[x1][y1][x2][y2][k] != -1) return dp[x1][y1][x2][y2][k];
    if (k == 1 || x1 == x2 || y1 == y2)
        return dp[x1][y1][x2][y2][k] = cal(x1, y1, x2, y2);
    int res = INF;
    for (a = x1; a < x2; ++ a) {
        res = min(res, min(solve(x1, y1, a, y2, k-1) + cal(a+1, y1, x2, y2), solve(a+1, y1, x2, y2, k-1) + cal(x1, y1, a, y2)));
    }
    for (b = y1; b < y2; ++ b) {
        res = min(res, min(solve(x1, y1, x2, b, k-1) + cal(x1, b+1, x2, y2), solve(x1, b+1, x2, y2, k-1) + cal(x1, y1, x2, b)));
    }
    dp[x1][y1][x2][y2][k] = res;
    return res;
}
int main() {
    #ifdef LOCAL
        freopen("data.in", "r", stdin);
    #endif // LOCAL
    int n;
    while (cin >> n) {
        memset(dp, -1, sizeof(dp));
        memset(mat, 0, sizeof(mat));
        double sum = 0.0;
        for (int i = 1; i <= MAXN; ++ i) {
            for (int j = 1; j <= MAXN; ++ j) {
                cin >> mat[i][j];
                sum += (double)mat[i][j];
            }
        }
        for (int i = 1; i <= MAXN; ++ i) {
            for (int j = 1; j <= MAXN; ++ j) {
                mat[i][j] = mat[i][j] + mat[i-1][j] + mat[i][j-1] - mat[i-1][j-1];
            }
        }
        sum /= (double)n; sum *= sum;
        solve(1, 1, MAXN, MAXN, n);
        double res = sqrt((double)dp[1][1][MAXN][MAXN][n]/n - sum);
        printf("%.3f\n", res);

    }
}