hdu 2830 Matrix Swapping II

最新推荐文章于 2018-11-18 19:28:50 发布

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#子矩阵

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本文介绍了一种通过动态规划和排序来求解给定0-1矩阵中最大全1子矩形面积的方法。该方法允许交换矩阵中的任意两列以增大子矩形的最大面积。

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 Given an N * M matrix with each entry equal to 0 or 1. We can find some rectangles in the matrix whose entries are all 1, and we define the maximum area of such rectangle as this matrix’s goodness.

We can swap any two columns any times, and we are to make the goodness of the matrix as large as possible.

Input
There are several test cases in the input. The first line of each test case contains two integers N and M (1 ≤ N,M ≤ 1000). Then N lines follow, each contains M numbers (0 or 1), indicating the N * M matrix
Output
Output one line for each test case, indicating the maximum possible goodness.
Sample Input

Sample Output

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这道题就是用dp[i][j]表示以i,j为底,包含当前位置的全有1组成的长度,由于可交换,我们对每一行的dp[i]进行排序之后直接遍历计算即可

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=1e3+10;
int arr[N][N];
int num[N];
char app[N][N];
int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    while(cin>>n>>m)
    {
        memset(arr,0,sizeof arr);
        for(int i=1;i<=n;i++)
            cin>>app[i];
        int mx=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(app[i][j]=='1')
                    arr[i][j]=arr[i-1][j]+1;
                else
                    arr[i][j]=0;
                num[j]=arr[i][j];
            }
            sort(num,num+m);
            for(int j=0;j<m;j++)
                mx=max(mx,num[j]*(m-j));
        }
        cout<<mx<<endl;
    }
    return 0;
}