Largest Submatrix of All 1’s（单调栈）

Largest Submatrix of All 1’s

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 9286		Accepted: 3336
Case Time Limit: 2000MS

Description

Given a m-by-n (0,1)-matrix, of all its submatrices of all 1’s which is the largest? By largest we mean that the submatrix has the most elements.

Input

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 2000) on line. Then come the elements of a (0,1)-matrix in row-major order on m lines each with n numbers. The input ends once EOF is met.

Output

For each test case, output one line containing the number of elements of the largest submatrix of all 1’s. If the given matrix is of all 0’s, output 0.

Sample Input

2 2
0 0
0 0
4 4
0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0

Sample Output

0
4

Source

POJ Founder Monthly Contest – 2008.01.31, xfxyjwf

 

解题思路： 枚举每一行 以改行为底矩形的最大值 转换为了这个问题(https://www.cnblogs.com/qq-1585047819/p/11347741.html); 维护两个单调队列

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;

int n,m,maxx;
const int N=2005;
int dp[N];
int sta[N],top;
int ma1[N],ma2[N];

inline int read(){
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    return x*f;
}

int main(){
    while(cin>>n>>m){
        memset(dp,0,sizeof(dp));
        memset(sta,0,sizeof(sta));
        top=0;maxx=0;
        for(int i=1;i<=n;i++){
            for(int j=1,d;j<=m;j++){
                d=read();
                dp[j]=d==0?0:++dp[j];
            }
            memset(sta,0,sizeof(sta));
            top=0;
            for(int j=1;j<=m;j++){
                int flag=0;
                while(dp[sta[top]]>=dp[j]&&top) top--,flag=1;
                if(flag==0) ma1[j]=j-1;
                else ma1[j]=top==0?0:sta[top];
                sta[++top]=j;
            }
            memset(sta,0,sizeof(sta));
            top=0;
            for(int j=m;j>=1;j--){
                int flag=0;
                while(dp[sta[top]]>=dp[j]&&top) top--,flag=1;
                if(flag==0) ma2[j]=j+1;
                else ma2[j]=top==0?m+1:sta[top];
                sta[++top]=j;
            }
            for(int j=1;j<=m;j++){
                maxx=max(maxx,(ma2[j]-ma1[j]-1)*dp[j]);
            }
        }
        cout << maxx << endl;
    }
}