H . Maximal submatrix（单调栈）

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,mn,m(1≤n,m≤2∗10^3)representing the size of the matrix 

the next nn line followed. the ii-th line contains mm integers vijvij(1≤vij≤5∗10^3) 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

Samples

Input Copy

1
2 3
1 2 4
2 3 3

Output

4



题目大意：给出一个N * M矩阵，求出最大的列上升子矩阵

for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(a[i][j]>=a[i-1][j]){
                    s[i][j]=s[i-1][j]+1;
                }
                else{
                    s[i][j]=1;
                }
            }
        }

然后对于这个题处理出来这个矩阵之后，就转化为这个题了：传送门

就是对于每一行对该矩阵求最大的子矩阵大小

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=5e3+100;
int a[maxn][maxn];
int s[maxn][maxn];
int h[maxn];
int q[maxn];
int l[maxn],r[maxn];
int n,m;
int judge(int h[]){
    h[0] = h[m + 1] = -1;
    int tt = -1;
    q[++ tt] = 0;
    for(int i = 1; i <= m; i ++)
    {
        while(h[q[tt]] >= h[i])  tt --;
        l[i] = q[tt]+1;
        q[++ tt] = i;
    }
    tt = -1;
    q[++ tt] = m + 1;
    for(int i = m; i; i --)
    {
        while(h[q[tt]] >= h[i])  tt --;
        r[i] = q[tt]-1;
        q[++ tt] = i;
    }
    int res = 0;
    for(int i = 1; i <= m; i ++)  res = max(res,h[i]*(r[i]-l[i]+1));
    return res;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        cin>>n>>m;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&a[i][j]);
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(a[i][j]>=a[i-1][j]){
                    s[i][j]=s[i-1][j]+1;
                }
                else{
                    s[i][j]=1;
                }
            }
        }
        int ans=0;
        for(int i=1;i<=n;i++){
            ans=max(ans,judge(s[i]));
        }
        cout<<ans<<endl;
    }
}