【递归 & 动态规划】POJ 1191 棋盘分割

大一的时候接触这道题目，当时死活不会做。。。最近突然又想起了这个伤疤，就把这道题给秒杀掉了，想想当年的我真是笨到家了T_T

首先对表达式进行变形，不考虑常量，最终其实就是求分成n块，使得每块的分值平方和最小。

基本思想很简单了，用dp[r1][c1][r2][c2]
表示r1行到r2行、c1列到c2列的矩形区域切n刀得到的最小平方和，n=0时是一个边界条件，对于n>0的情形，分纵切和横切两种情况，对每一种情况，对左（上）半部分或者右（下）半部分继续切n-1刀，取四种情况的最小值就可以了。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cassert>
#include <cmath>
#include <limits.h>
#define FOR(i, n) for (int i = 0; i < n; ++i)
using namespace std;

int arr[9][9], sm[9][9];
int dp[9][9][9][9][16];
// get the square of sum in this region
int get(int r1, int c1, int r2, int c2) {
assert(r1 <= r2 && c1 <= c2);
int res = sm[r2][c2] - sm[r2][c1 - 1] - sm[r1 - 1][c2] + sm[r1 - 1][c1 - 1];
return res * res;
}

const int INF = INT_MAX / 4;
int solve(int r1, int c1, int r2, int c2, int n) {
assert(n >= 0);
if (dp[r1][c1][r2][c2]
!= -1) return dp[r1][c1][r2][c2]
;
if (r1 > r2 || c1 > c2) return INF;
if (n == 0) return get(r1, c1, r2, c2);
// can not be divided anymore
if (r1 == r2 && c1 == c2) return INF;
int mn = INF;
// split in column
for (int c = c1; c < c2; ++c) {
// continue to split on right part
mn = min(mn, get(r1, c1, r2, c) + solve(r1, c + 1, r2, c2, n - 1));
// continue to split on left part
mn = min(mn, solve(r1, c1, r2, c, n - 1) + get(r1, c + 1, r2, c2));
}
// split in row
for (int r = r1; r < r2; ++r) {
// continue to split on bottom part
mn = min(mn, get(r1, c1, r, c2) + solve(r + 1, c1, r2, c2, n - 1));
// continue to split on top part
mn = min(mn, solve(r1, c1, r, c2, n - 1) + get(r + 1, c1, r2, c2));
}
dp[r1][c1][r2][c2]
= mn;
return mn;
}

int main() {
int N, total = 0;
scanf("%d", &N);
memset(arr, 0, sizeof(arr));
memset(sm, 0, sizeof(sm));
FOR(i, 8) FOR(j, 8) {
scanf("%d", arr[i + 1] + j + 1);
total += arr[i + 1][j + 1];
}
for (int i = 1; i <= 8; ++i)
for (int j = 1; j <= 8; ++j)
sm[i][j] = sm[i - 1][j] + sm[i][j - 1] - sm[i - 1][j - 1] + arr[i][j];
memset(dp, -1, sizeof(dp));
double res = solve(1, 1, 8, 8, N - 1) - static_cast<double>(total * total) / N;
res = sqrt(res / N);
// answer for sample: 1.633
printf("%.3f\n", res);
return 0;
}