Matrix Swapping II
Time Limit : 9000/3000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 26 Accepted Submission(s) : 17
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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
Sample Output
4 2 Note: Huge Input, scanf() is recommended.
Source
2009 Multi-University Training Contest 2 - Host by TJU
//for(int j=n-1;j>=1;j--) dp[j]+=dp[j+1];
/*
1011
1111
1101
1101
如果没有这句话答案是8,有了这句正确的是9,就是有可能下半部分要,不要上半部分的1会更大
*/
#include <bits/stdc++.h>
using namespace std;
char a;
int dp[1005],h[1005][1005],n,m; //h[i][j]代表包括到第i行第j列,前面有几个连续的1
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
getchar(); //很重要,不能少
memset(h,0,sizeof(h));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf("%c",&a);
if(a=='1') h[i][j]=h[i-1][j]+1; //如果为1,就是上一行的数+1
else h[i][j]=0; //否则清0
}
getchar(); //接受换行符,否则读入就会出错
}
int ans=-1;
for(int i=1;i<=n;i++)
{
memset(dp,0,sizeof(dp));
for(int j=1;j<=m;j++) //每一行中1、2、3……有几个
dp[h[i][j]]++;
for(int j=n-1;j>=1;j--) //!!!!!!少了就wa
dp[j]+=dp[j+1];
for(int j=1;j<=n;j++) //更新答案
ans=max(ans,dp[j]*j);
}
printf("%d\n",ans);
}
return 0;
}