每一行判断高度,然后排序,枚举计算最大面积
Matrix Swapping II
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 896 Accepted Submission(s): 597
Problem Description
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.
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
3 4
1011
1001
0001
3 4
1010
1001
0001
1011
1001
0001
3 4
1010
1001
0001
Sample Output
4
2
Note: Huge Input, scanf() is recommended.
2
Note: Huge Input, scanf() is recommended.
Source
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=1005;
struct m
{
char a[N][N];
int high[N];
}G;
int main()
{
int R,C;
while(scanf("%d%d",&R,&C)!=EOF)
{
int ans=-99999999;
for(int i=0;i<R;i++)
scanf("%s",G.a);
for(int i=0;i<R;i++)
{
memset(G.high,0,sizeof(G.high));
for(int j=0;j<C;j++)
{
for(int k=i;k>=0;k--)
{
if(G.a[k][j]=='1')
G.high[j]++;
else break;
}
}
sort(G.high,G.high+C);
/*
cout<<i<<":"<<endl;
for(int j=0;j<C;j++)
cout<<G.high[j]<<" ";
cout<<endl;
*/
for(int i=0;i<C;i++)
{
int cnt=C-i;
// cout<<cnt<<"*"<<G.high<<endl;
ans=max(ans,cnt*G.high);
}
}
printf("%d\n",ans);
}
return 0;
}