2019牛客多校第三场F Planting Trees（单调队列）题解

题意：

求最大矩阵面积，要求矩阵内数字满足\(max - min < m\)

思路：

枚举上下长度，在枚举的时候可以求出每一列的最大最小值\(cmax，cmin\)，这样问题就变成了求一行数，要你得到\(max - min < m\)的最长长度。用单调队列\(O(n)\)求解。总复杂度\(O(n^3)\)。

代码：

#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<stack>
#include<ctime>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn = 500 + 5;
const int INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;
int n, m;
int a[maxn][maxn];
int cmax[maxn], cmin[maxn];  //递减 递增
int maxq[maxn], minq[maxn];
int getans(){
    int maxhead = 0, maxtail = 0, minhead = 0, mintail = 0;
    int l = 1;
    int len = 0;
    for(int i = 1; i <= n; i++){
        while(maxhead < maxtail && cmax[i] > cmax[maxq[maxtail - 1]]) maxtail--;
        maxq[maxtail++] = i;
        while(minhead < mintail && cmin[i] < cmin[minq[mintail - 1]]) mintail--;
        minq[mintail++] = i;

        while(minhead < mintail && maxhead < maxtail && cmax[maxq[maxhead]] - cmin[minq[minhead]] > m){
            l = min(maxq[maxhead], minq[minhead]) + 1;  //跳到l + 1
            while(minhead < mintail && minq[minhead] < l) minhead++;
            while(maxhead < maxtail && maxq[maxhead] < l) maxhead++;
        }

        len = max(len, i - l + 1);
    }
    return len;
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                scanf("%d", &a[i][j]);
            }
        }
        int ans = 0;
        for(int i = 1; i <= n; i++){    //从i
            for(int j = 1; j <= n; j++){
                cmax[j] = -1;
                cmin[j] = INF;
            }
            for(int j = i; j <= n; j++){    //到j
                for(int k = 1; k <= n; k++){
                    cmax[k] = max(cmax[k], a[j][k]);
                    cmin[k] = min(cmin[k], a[j][k]);
                }
                ans = max(ans, (j - i + 1) * getans());

            }
        }
        printf("%d\n", ans);
    }

    return 0;
}