7.20 第一场 Maximal submatrix

Maximal submatrix

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 276 Accepted Submission(s): 76

Problem Description

Given a matrix of n rows and m columns,find the largest area submatrix which is non decreasing on each column

Input

The first line contains an integer T(1≤T≤10)representing the number of test cases.
For each test case, the first line contains two integers n,m(1≤n,m≤2∗10³)representing the size of the matrix
the next n line followed. the i-th line contains m integers v_ij(1≤v_ij≤5∗10³)representing the value of matrix
It is guaranteed that there are no more than 2 testcases with n∗m>10000

Output

For each test case, print a integer representing the Maximal submatrix

Sample Input

1
2 3
1 2 4
2 3 3

Sample Output

4

大概题意

给出一个矩阵，求这个矩阵列上递增的最大子矩阵

思路

遍历矩阵，求每个点所在列递增连续元素个数，然后从底向上遍历矩阵，找到当前行每个元素所在列连续递增元素个数，取最小连续递增元素个数与列数之积即为最大子矩阵。

代码

//
// Created by Black on 2021/7/20.
//
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 3010;
int n, m, t, a[N][N], up[N][N], ans, res;

int main() {
//    cin >> t;
    scanf("%d", &t);
    while (t--) {
        memset(up, 0, sizeof up);
//        cin >> n >> m;
        scanf("%d%d", &n, &m);
//        if (n == 1) {
//            cout << m << endl;
//            continue;
//        }
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                scanf("%d", &a[i][j]);

        for (int i = 2; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (a[i][j] >= a[i - 1][j]) {
                    up[i][j] = up[i - 1][j] + 1;
                }
            }
        }
        ans = 0;
        for (int i = n; i > 0; i--) {
            int last = 0;
            res = 10000;
            for (int j = 1; j <= m; j++) {
                if (up[i][j])
                    res = min(res, up[i][j] + 1);
                else {
                    ans = max(ans, res * (j - last - 1));
                    last = j;
                }
            }
            if (res == 10000)
                continue;
            ans = max(ans, res * (m - last));
        }
        if (ans < m)
            printf("%d\n", m);
        else
            printf("%d\n", ans);
    }
    return 0;
}