poj1191 棋盤分割 dp

本題黑書P116有詳細講解。

思路：

先把方差公式化簡：

σ2=1n∑i=1nx2i−(x¯)2

其中平均數是一定的，就是方格里所有的數的和除以n，所以問題轉化爲將棋盤分成n個矩形，使每個矩形的總分的平方和最小。

考慮左上角爲（x1，y1），右下角爲（x2，y2）的矩形：
設它的總分爲s[x1, y1, x2, y2]，切割k次得到k+1塊矩形的總分平方和最小爲d[x1, y1, x2, y2]，則切一刀時，可以橫着切，也可以豎着切，然後遞歸下去繼續切即可。

代碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <climits>

#define THIS (dp[k][x1][y1][x2][y2])

using namespace std;

int n;
int arr[11][11];
double sum[11][11];
double dp[17][11][11][11][11];

inline double S(int x1, int y1, int x2, int y2) {
    double t = sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
    return t * t;
}

double dfs(int k, int x1, int y1, int x2, int y2) {
    if (k == 1) {
        THIS = S(x1, y1, x2, y2);
        return THIS;
    }
    if (fabs(THIS) > 1e-8) return THIS;
    THIS = 1 << 29;
    for (int i = x1; i < x2; i++) {
        THIS = min(THIS, min(dfs(k - 1, x1, y1, i, y2) + S(i + 1, y1, x2, y2),
            dfs(k - 1, i + 1, y1, x2, y2) + S(x1, y1, i, y2)));
    }
    for (int i = y1; i < y2; i++) {
        THIS = min(THIS, min(dfs(k - 1, x1, y1, x2, i) + S(x1, i + 1, x2, y2),
            dfs(k - 1, x1, i + 1, x2, y2) + S(x1, y1, x2, i)));
    }
    return THIS;
}

int main() {
    freopen("1191.in", "r", stdin);
    cin >> n;
    for (int i = 1; i <= 8; i++)
        for (int j = 1; j <= 8; j++)
            cin >> arr[i][j];
    for (int i = 1; i <= 8; i++)
        for (int j = 1; j <= 8; j++)
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + arr[i][j];
    double adv = sum[8][8] / n;
    double ans = dfs(n, 1, 1, 8, 8) / n;
    printf("%.3lf\n", sqrt(ans - adv * adv));
    return 0;
}