理想正方形

https://www.acwing.com/problem/content/1093/

\(有一个 a×b 的整数组成的矩阵，现请你从中找出一个 n×n 的正方形区域，使得该区域所有数中的最大值和最小值的差最小。\)

思路

\(分别得到每个n*n正方形中的最小值和最大值\)

• \(对于每一行做一遍单调队列得到a*(b-n+1)的矩阵,\\ 再对于每一列,竖着做一遍单调队列,得到(a-n+1)*(b-n+1)的矩阵,每个元素代表n*n的最大值(最小值)\)

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define fi first
#define se second
#define inf 0x3f3f3f3f
const int N = 1010;
int w[N][N], t1[N], t2[N], t3[N];
int n, m, k;
int ans_max[N][N], ans_min[N][N];
int q[N];

void get_max(int a[], int ans[], int m) {
    int hh = 0, tt = -1, cnt = 0;
    for (int i = 1; i <= m; ++i) {
        if (hh <= tt && q[hh] <= i - k) ++hh;
        while (hh <= tt && a[q[tt]] <= a[i]) --tt;
        q[++tt] = i;
        if (i >= k) ans[++cnt] = a[q[hh]];
    }
}

void get_min(int a[], int ans[], int m) {
    int hh = 0, tt = -1, cnt = 0;
    for (int i = 1; i <= m; ++i) {
        if (hh <= tt && q[hh] <= i - k) ++hh;
        while (hh <= tt && a[q[tt]] >= a[i]) --tt;
        q[++tt] = i;
        if (i >= k) ans[++cnt] = a[q[hh]];
    }
}
           
int main() {
    IO;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            cin >> w[i][j];
    for (int i = 1; i <= n; ++i) {
        get_max(w[i], ans_max[i], m);
        get_min(w[i], ans_min[i], m);
    }
    for (int j = 1; j <= m - k + 1; ++j) {
        for (int i = 1; i <= n; ++i) {
            t1[i] = ans_max[i][j];
            t2[i] = ans_min[i][j];
        }
        get_max(t1, t3, n);
        for (int i = 1; i <= n - k + 1; ++i) ans_max[i][j] = t3[i];
        get_min(t2, t3, n);
        for (int i = 1; i <= n - k + 1; ++i) ans_min[i][j] = t3[i];
    }
    int ans = inf;
    for (int i = 1; i <= n - k + 1; ++i)
        for (int j = 1; j <= m - k + 1; ++j)
            ans = min(ans, ans_max[i][j] - ans_min[i][j]);
    cout << ans << '\n';
    return 0;
}