POJ 1191 棋盘分割（记忆化搜索）

题意：

现在需要把棋盘分割成 n 块矩形棋盘，并使各矩形棋盘总分的均方差最小。

思路：

化简公式，记忆化搜索。

 

#include <iostream>

#include <algorithm>

#include <cmath>

using namespace std;



const int INFS = 0x3FFFFFFF;

int square[10][10][10][10], grid[10][10];

int dp[10][10][10][10][16];



void initsquare() {

    for (int i = 1; i <= 8; i++) {

        for (int j = 1; j <= 8; j++) {

            for (int p = i; p <= 8; p++) {

                for (int q = j; q <= 8; q++) {

                    int t = 0;

                    for (int x = i; x <= p; x++)

                        for (int y = j; y <= q; y++)

                            t += grid[x][y];

                    square[i][j][p][q] = t * t;

                }

            }

        }

    }

}



int solvedp(int k, int x, int y, int ex, int ey) {

    if (k == 1)

        return square[x][y][ex][ey];



    if (dp[x][y][ex][ey][k] != -1)

        return dp[x][y][ex][ey][k];



    int ans = INFS;

    for (int i = x; i < ex; i++) {

        ans = min(ans, square[x][y][i][ey] + solvedp(k-1, i+1, y, ex, ey));

        ans = min(ans, square[i+1][y][ex][ey] + solvedp(k-1, x, y, i, ey));

    } 

    for (int i = y; i < ey; i++) {

        ans = min(ans, square[x][y][ex][i] + solvedp(k-1, x, i+1, ex, ey));

        ans = min(ans, square[x][i+1][ex][ey] + solvedp(k-1, x, y, ex, i));

    }

    return dp[x][y][ex][ey][k] = ans;

}



int main() {

    int n;

    scanf("%d", &n);



    double ave = 0;

    for (int i = 1; i <= 8; i++) {

        for (int j = 1; j <= 8; j++) {

            scanf("%d", &grid[i][j]);

            ave += grid[i][j];

        }

    }

    initsquare();

    memset(dp, -1, sizeof(dp));

    double ans = solvedp(n, 1, 1, 8, 8);

    ave /= n;

    ans /= n;

    ans -= ave * ave;

    printf("%.3lf\n", sqrt(ans));

    return 0;

}