【题解】牛客 Planting Trees⭐⭐⭐ 【单调队列】

牛客 Planting Trees⭐⭐⭐

Input

Output

For each case, print a single integer, the maximum number of cells in a valid rectangle.

Examples

2
2 0
1 2
2 1
3 1
1 3 2
2 3 1
3 2 1
输出
复制
1
4

Hint

题意:

给出一个矩阵, 找到一个子矩阵满足矩阵内任意2元素差不超过M, 求子矩阵的最打的积

题解:

枚举上下边界, 同时枚举左边界, 通过单调队列可以确定最大满足题意的右边界

经验小结:

#include
using namespace std;
#define ms(x, n) memset(x,n,sizeof(x));
typedef  long long LL;
const int INF = 1 << 30;
const int MAXN = 510;

int T, n, m, a[MAXN][MAXN];
int cmax[MAXN], cmin[MAXN]; //每一列的最大值/最小值
int qmin[MAXN], qmax[MAXN], head1, head2, tail1, tail2;
int getAns(){
    head1 = head2 = tail1 = tail2 = 0;
    int l = 1, ret = 0;
    for(int r = 1; r <= n; ++r){ //枚举右边界, 通过单调队列确定左边界
        while(head1 < tail1 && cmin[qmin[tail1-1]] > cmin[r])
            --tail1;
        qmin[tail1++] = r;
        while(head2 < tail2 && cmax[qmax[tail2-1]] < cmax[r])
            --tail2;
        qmax[tail2++] = r;
        while(head1 < tail1 && head2 < tail2 && cmax[qmax[head2]]-cmin[qmin[head1]] > m){
            l = min(qmax[head2], qmin[head1]) + 1;
            while(head1 < tail1 && qmin[head1] < l)
                ++head1;
            while(head2 < tail2 && qmax[head2] < l)
                ++head2;
        }
        ret = max(ret, r-l+1);
    }
    return ret;
}
int main() {
    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){ //i, j表示枚举上下界
            for(int j = 1; j <= n; ++j)
                cmax[j] = -1, cmin[j] = INF;
            for(int j = i; j <= n; ++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;
}