POJ3494Largest Submatrix of All 1’s（DP）

最新推荐文章于 2020-08-12 20:21:32 发布

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最新推荐文章于 2020-08-12 20:21:32 发布

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分类专栏： ACM 文章标签： output input each matrix

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本文介绍了一个算法问题：给定一个由0和1组成的矩阵，如何找出全是1的最大子矩阵，并给出了具体的输入输出样例及解决方案实现代码。

Largest Submatrix of All 1’s

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 2740		Accepted: 971
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

这个题目我不想多说，和2599一样。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int left1[2001],right1[2001],h[2001],n,m; 
char s[4020];
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    //while(cin>>n>>m)
    {
         int i,j,k,ans=0,temp;
         //memset(left,0,sizeof(left));
         //memset(right,0,sizeof(right));
         memset(h,0,sizeof(h));
         for(i=1;i<=m;i++)  
         {
              left1[i]=i;right1[i]=i;
         }
         getchar();
         for(i=1;i<=n;i++)
         {
               //for(j=1;j<=m;j++) 
               //{
                     //scanf("%d",&s);
               gets(s);
               for(j=1;j<=m;j++)
               {
                     if(s[2*j-2]=='1')  h[j]++;
                     else{  h[j]=0;}
                     left1[j]=j;right1[j]=j;
               } 
               //}
               for(j=2;j<=m;j++)
               {
                    if(h[j]==0)  continue;
                    for(k=j-1;k>0;) 
                    {
                        if(h[j]<=h[k])
                        {
                              left1[j]=left1[k];k=left1[k]-1;
                        }
                        else break;
                    }
               } 
               for(j=m-1;j>0;j--)
               {
                    if(h[j]==0)  continue;
                    for(k=j+1;k<=m;) 
                    {
                        if(h[j]<=h[k])
                        {
                              right1[j]=right1[k];k=right1[k]+1;
                        }
                        else break;         
                    }
               } 
               for(j=1;j<=m;j++)
               {
                    if(h[j]==0) continue;
                    temp=(right1[j]-left1[j]+1)*h[j];
                    if(temp>ans)  ans=temp;
               }
         } 
         cout<<ans<<endl;
         //printf("%d/n",ans);
    }
    return 0;
}