BZOJ 1047: [HAOI2007]理想的正方形

1047

思路：

二维单调队列

先用单调队列预处理出每个位置在这一行往前n个位置的最大值最小值，然后O(n^2)扫过去，对列用单调队列

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 1e3 + 5;
const int INF = 0x7f7f7f7f;
int num[N][N], mx[N][N], mn[N][N];
int main() {
    int a, b, n;
    scanf("%d %d %d", &a, &b, &n);
    for (int i = 1; i <= a; i++) {
        for (int j = 1; j <= b; j++) {
            scanf("%d", &num[i][j]);
        }
    }
    deque<int>q1, q2;
    for (int i = 1; i <= a; i++) {
        q1.clear();
        q2.clear();
        for (int j = 1; j <= b; j++) {
            while(!q1.empty() && num[i][q1.back()] >= num[i][j]) q1.pop_back();
            q1.push_back(j);
            while(!q1.empty() && q1.front() < j - n + 1) q1.pop_front();
            mn[i][j] = num[i][q1.front()];

            while(!q2.empty() && num[i][q2.back()] <= num[i][j]) q2.pop_back();
            q2.push_back(j);
            while(!q2.empty() && q2.front() < j - n + 1) q2.pop_front();
            mx[i][j] = num[i][q2.front()];
        }
    }
    int ans = INF;
    for (int j = n; j <= b; j++) {
        q1.clear();
        q2.clear();
        for (int i = 1; i <= a; i++) {
            while(!q1.empty() && mn[q1.back()][j] >= mn[i][j]) q1.pop_back();
            q1.push_back(i);
            while(!q1.empty() && q1.front() < i - n + 1) q1.pop_front();

            while(!q2.empty() && mx[q2.back()][j] <= mx[i][j]) q2.pop_back();
            q2.push_back(i);
            while(!q2.empty() && q2.front() < i - n + 1) q2.pop_front();

            if(i >= n) {
                ans = min(ans, mx[q2.front()][j] - mn[q1.front()][j]);
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}