Monotonous stack (the largest sub-rectangular enhanced version) - Cattle off more school eighth field A

01 demand matrix in the number of different matrix 1

First, the pre-pre [i] [j] denotes the number of consecutive i 1 above, the height of each row of stack processing monotone

The stack elements maintain two values: pre [i] [j] and a forwardly extending up to maintain the position pos

Then calculate contributions, maintain a rightmost column of Table 1 below no position p sweep from left to right,

It can be determined whether the element includes a p-pos, when the pop-up, if this can be described is a matrix element represents contributes

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
typedef vector<int>vi;
 
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define fi first
#define se second
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" " ;
#define pb(x) push_back(x)
#define per(i,a,b) for(int i=(b)-1;i>=(a);--i)
const int N=5e3+5;
char mp[N][N];
int h[N][N];
int area[N][N];
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;++i)
        scanf("%s",mp[i]+1);
    
    memset(h,0,sizeof(h));
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)
            if(mp[i][j]=='1'){
                h[i][j]=h[i-1][j]+1;
            }
    int ans=0;
    for(int i=1;i<=n;++i){
        stack<pii>stk;
        int ma=-1;
        for(int j=1;j<=m+1;++j){
            int pos=j;
            while(!stk.empty()&&stk.top().fi>h[i][j]){
cout<<stk.top().fi<<'\n';
                if(stk.top().se<=ma){
                    ans++;
                }
                pos=stk.top().se;
                stk.pop();
            }
            if(!h[i+1][j])ma=j;
            if(h[i][j]&& ( stk.empty() || (stk.top().fi<h[i][j]) ) )
                stk.push(pii(h[i][j],pos));
        }
    }
    cout<<ans<<endl;
    return 0;
}