POJ - 3494 Largest Submatrix of All 1’s（单调栈）

最新推荐文章于 2019-07-15 14:53:55 发布

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最新推荐文章于 2019-07-15 14:53:55 发布

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本文介绍了一种使用单调栈解决最大全1子矩阵问题的方法。通过将矩阵中的每个1视为不同高度的矩形，利用单调栈计算这些矩形组成的最大面积子矩阵。文章提供了一个详细的C++实现示例。

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题目链接：http://poj.org/problem?id=3494点击打开链接

Largest Submatrix of All 1’s

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 6674		Accepted: 2472
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

Sample Output

0
4

单调栈的题将矩阵里的每个1从上往下看做是一个长方形则长度比他高的可以继续比他小的停止

#include <iostream>
#include <queue>
#include <stdio.h>
#include <stdlib.h>
#include <stack>
#include <limits>
#include <string>
#include <string.h>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
using namespace std;
int mmap[2222][2222];
int step[2222][2222];
int l[2222][2222];
int r[2222][2222];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(mmap,0,sizeof(mmap));
        memset(step,0,sizeof(mmap));
        memset(l,0,sizeof(mmap));
        memset(r,0,sizeof(mmap));
        for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=m;j++)
                    scanf("%d",&mmap[i][j]);
            }

        int len=1;
        for(int j=1;j<=m;j++)
            {
                for(int i=1;i<=n;i++)
                {
                    if(mmap[i][j])
                    {
                        step[i][j]=len++;
                    }
                    else
                    {
                        len=1;
                    }
                }
                len=1;
            }
        stack<int > s;
        for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=m;j++)
                   {
                            while(1)
                            {
                                if(s.empty())
                                    {
                                        if(mmap[i][j]==0)
                                            l[i][j]=0;
                                        else
                                            l[i][j]=1;
                                        s.push(j);
                                        break;
                                    }
                                    else
                                    {
                                        if(step[i][s.top()]>=step[i][j])
                                        {
                                            l[i][j]=s.top();
                                            s.pop();
                                        }
                                        else
                                        {
                                            l[i][j]=s.top()+1;
                                            s.push(j);
                                            break;
                                        }

                                    }
                            }
                   }
                   while(!s.empty())
                            s.pop();
            }

            for(int i=1;i<=n;i++)
            {
                for(int j=m;j>=1;j--)
                   {
                            while(1)
                            {
                                if(s.empty())
                                    {
                                        if(mmap[i][j]==0)
                                            r[i][j]=0;
                                        else
                                            r[i][j]=m;
                                        s.push(j);
                                        break;
                                    }
                                    else
                                    {
                                        if(step[i][s.top()]>=step[i][j])
                                        {
                                            r[i][j]=s.top();
                                            s.pop();
                                        }
                                        else
                                        {
                                            r[i][j]=s.top()-1;
                                            s.push(j);
                                            break;
                                        }

                                    }
                            }
                   }
                   while(!s.empty())
                            s.pop();
            }
            int mmax=0;
            for(int i=1;i<=n;i++)
                for(int j=1;j<=m;j++)
            {
                mmax=max(mmax,(r[i][j]-l[i][j]+1)*step[i][j]);
            }
             printf("%d\n",mmax);
            }
}