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[AcWing 321] 棋盘分割

作者:互联网

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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;
}

  1. 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
  2. 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

标签:y2,int,double,x1,x2,321,y1,棋盘,AcWing
来源: https://www.cnblogs.com/wKingYu/p/16461660.html