遞迴-棋盤分割

問題講解：

原始碼:

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;

int s[9][9];  //每個格子的分數
int sum[9][9];   //從(1，1)到(i，j)的矩形的分數之和
int res[15][9][9][9][9];     //fun的記錄表

int calSum(int x1, int y1, int x2, int y2) {    //(x1, y1)到(x2, y2)的矩形的分數之和
 

    return sum[x2][y2]-sum[x2][y1-1]-sum[x1-1][y2]+sum[x1-1][y1-1];
}

int fun(int n, int x1, int y1, int x2, int y2) {
    int t, a, b, c, e, MIN=10000000;
    if(res[n][x1][y1][x2][y2] != -1)
        return res[n][x1][y1][x2][y2];
    if (n==1) {
        t=calSum(x1, y1, x2, y2);
        res[n][x1][y1][x2][y2] = t*t;
        return
 
 t*t;
    }
    for (a=x1; a<x2; a++) {
        c=calSum(a+1, y1, x2, y2);
        e=calSum(x1, y1, a, y2);
        t=min(fun(n-1, x1, y1, a, y2)+c*c, fun(n-1, a+1, y1, x2, y2)+e*e);
        if(MIN>t) MIN=t;
    }
    for (b=y1; b<y2; b++) {
        c=calSum(x1, b+1, x2, y2);
        e=calSum(x1, y1, x2, b);
        t=min(fun(n-1
 
, x1, y1, x2, b)+c*c, fun(n-1, x1, b+1, x2, y2)+e*e);
        if(MIN>t) MIN=t;
    }
    res[n][x1][y1][x2][y2]=MIN;
    return MIN;
}

int main() {
    memset(sum, 0, sizeof(sum));
    memset(res, -1, sizeof(res));  //初始化記錄表
    int n;
    cin>>n;
    for(int i=1; i<9; i++) {
        for (int j=1, rowsum=0; j<9; j++) {
            cin >> s[i][j];
            rowsum += s[i][j];
            sum[i][j] += sum[i-1][j]+rowsum;
        }
    }
    double result = n*fun(n, 1, 1, 8, 8)-sum[8][8]*sum[8][8];
    cout << setiosflags(ios::fixed) << setprecision(3) << sqrt(result/(n*n)) << endl;
    return 0;
}