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#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long LL;
const int N = 10, M = 20;
const double INF = 1e9;
int n, m = 8;
int s[N][N];
double f[N][N][N][N][M];
double xx;
int front_sum(int x1, int y1, int x2, int y2)
{
return s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
}
double get(int x1, int y1, int x2, int y2)
{
double sum = front_sum(x1, y1, x2, y2) - xx;
return sum * sum / n;
}
double dp(int x1, int y1, int x2, int y2, int k)
{
double &v = f[x1][y1][x2][y2][k];
if (v >= 0)
return v;
if (k == 1)
return v = get(x1, y1, x2, y2);
v = INF;
for (int i = x1; i < x2; i ++) {
v = min(v, dp(x1, y1, i, y2, k - 1) + get(i + 1, y1, x2, y2));
v = min(v, dp(i + 1, y1, x2, y2, k - 1) + get(x1, y1, i, y2));
}
for (int i = y1; i < y2; i ++) {
v = min(v, dp(x1, y1, x2, i, k - 1) + get(x1, i + 1, x2, y2));
v = min(v, dp(x1, i + 1, x2, y2, k - 1) + get(x1, y1, x2, i));
}
return v;
}
int main()
{
cin >> n;
for (int i = 1; i <= m; i ++)
for (int j = 1; j <= m; j ++) {
cin >> s[i][j];
s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
}
xx = (double)s[m][m] / n;
memset(f, -1, sizeof f);
printf("%.3f/n", sqrt(dp(1, 1, 8, 8, n)));
return 0;
}
- 状态表示
/(f[x_1][y_1][x_2][y_2][k]/) 表示所有将矩阵 /((x_1, y_1),(x_2, y_2)/) 分割为 /(k/) 份的均方差的平方的最小值 - 状态计算
考虑将矩阵 /((x_1, y_1),(x_2, y_2)/) 分割为 /(2/) 份的情况,也就是在矩阵中划一条分割线,将矩阵分为两部分,考虑分割线水平和分割线竖直分两种情况:
① 分割线水平,分割线位置的取值为 /(/lambda / /epsilon / [y_1, y_2)/),划分为 /([x_1, y_1, x_2, /lambda]/) 和 /([x_1, /lambda + 1, x_2, y_2]/) 两部分
① 分割线竖直,分割线位置的取值为 /(/lambda / /epsilon / [x_1, x_2)/),划分为 /([x_1, y_1, /lambda, y_2]/) 和 /([/lambda + 1, y1, x_2, y_2]/) 两部分
原创文章,作者:kepupublish,如若转载,请注明出处:https://blog.ytso.com/273025.html