# 【BZOJ1048】[HAOI2007]分割矩阵

bzoj

## 题解

$$dp[a][b][c][d][num]$$表示将矩形$$(a,b,c,d)$$分成$$num$$个的最小方差，然后转移即可

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
#define sqr(x) ((x) * (x))
int A, B, N, a[15][15], s[15][15];
double ave, dp[15][15][15][15][15];
int sum(int x, int y, int _x, int _y) { return s[_x][_y] - s[_x][y - 1] - s[x - 1][_y] + s[x - 1][y - 1]; }
double dfs(int x, int y, int _x, int _y, int num) {
if (dp[x][y][_x][_y][num] != 0) return dp[x][y][_x][_y][num];
else if (num == 1) return 1.0 * sqr(sum(x, y, _x, _y) - ave);
dp[x][y][_x][_y][num] = 1e9;
for (int i = y; i < _y; i++) {
for (int j = 1; j < num; j++) {
double res1 = dfs(x, y, _x, i, j) + dfs(x, i + 1, _x, _y, num - j);
double res2 = dfs(x, i + 1, _x, _y, j) + dfs(x, y, _x, i, num - j);
dp[x][y][_x][_y][num] = min(dp[x][y][_x][_y][num], min(res1, res2));
}
}
for (int i = x; i < _x; i++) {
for (int j = 1; j < num; j++) {
double res1 = dfs(x, y, i, _y, j) + dfs(i + 1, y, _x, _y, num - j);
double res2 = dfs(i + 1, y, _x, _y, j) + dfs(x, y, i, _y, num - j);
dp[x][y][_x][_y][num] = min(dp[x][y][_x][_y][num], min(res1, res2));
}
}
return dp[x][y][_x][_y][num];
}
int main () {
cin >> A >> B >> N;
for (int i = 1; i <= A; i++)
for (int j = 1; j <= B; j++) cin >> a[i][j];
for (int i = 1; i <= A; i++)
for (int j = 1; j <= B; j++)
s[i][j] = s[i][j - 1] + s[i - 1][j] - s[i - 1][j - 1] + a[i][j];
ave = 1.0 * s[A][B] / N;
printf("%0.2lf\n", sqrt(dfs(1, 1, A, B, N) / N));
return 0;
}

posted @ 2018-12-26 11:56  heyujun  阅读(260)  评论(0编辑  收藏  举报