All-one Matrices

题目链接：https://ac.nowcoder.com/acm/contest/888/A

题目描述

Gromah and LZR entered the great tomb, the first thing they see is a matrix of size n×mn\times mn×m, and the elements in the matrix are all 0 or 1​.

LZR finds a note board saying "An all-one matrix is defined as the matrix whose elements are all 1​, you should determine the number of all-one submatrices of the given matrix that are not completely included by any other all-one submatrices".

Meanwhile, Gromah also finds a password lock, obviously the password should be the number mentioned in the note board!

Please help them determine the password and enter the next level.

输入描述:

The first line contains two positive integers n,m​, denoting the size of given matrix.

 

Following n lines each contains a string with length m, whose elements are all 0 or 1​, denoting the given matrix.

 

 

1≤n,m≤3000

输出描述:

Print a non-negative integer, denoting the answer.

示例1

输入

复制

3 4
0111
1110
0101

输出

复制

5

说明

The 5 matrices are (1,2)−(1,4),  (1,2)−(2,3),  (1,2)−(3,2),  (2,1)−(2,3),  (3,4)−(3,4)

#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 3005
int a[maxn][maxn];
int h[maxn];//存该列往上有多少个1
int l[maxn],r[maxn];
int vis[maxn][maxn];
struct node{
    int L,R;
}hv[maxn][maxn];
int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    int ans=0;
    cin>>n>>m;
    char ch;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>ch;
            a[i][j]=ch=='0'?0:1;
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]==1) h[j]++;
            else h[j]=0;
        }
        h[0]=h[m+1]=-1;
        for(int j=1;j<=m;j++)//往左最远走到哪 
        {
            int k=j;
            while(h[j]<=h[k-1]) k=l[k-1];
            l[j]=k;
        }
        for(int j=m;j>=1;j--)//往右最远走到哪
        {
            int k=j;
            while(h[j]<=h[k+1]) k=r[k+1];
            r[j]=k;
        }
        for(int j=1;j<=m;j++)
        {
            if(h[j]&&!vis[l[j]][r[j]]&&(hv[i-1][j].L!=l[j]||hv[i-1][j].R!=r[j]))
            {
                ans++;
                vis[l[j]][r[j]]=1;
            }
            if(h[j])
            {
                hv[i][j].L=l[j];
                hv[i][j].R=r[j];
            }
        }
        for(int j=1;j<=m;j++)
        vis[l[j]][r[j]]=0;
    }
    cout<<ans<<endl;
    return 0;
}