Matrix Swapping II

Matrix Swapping II

Time Limit: 9000/3000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 906Accepted Submission(s): 602

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.

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.

和前面的求最大子1矩阵类似。只是这部不用求左右界限了，而是求出每个高度出现的次数

#include <stdio.h>
#include <string.h>
int m, n, h[1001], num[1001];
char matrix[1001];
int main() {
	//freopen("in.txt", "r", stdin);
	while(scanf("%d %d",&m, &n) != EOF) {
		getchar();
		memset(h,0,sizeof(int)*n);
		int max = 0;
		for(int i=1; i<=m; i++) {
			gets(matrix);
			memset(num,0,sizeof(int)*(i+1));
			for(int j=0; j<n; j++) {
				if(matrix[j] == '1')
				h[j]++;
				else
				h[j] = 0;
				num[h[j]]++;
				for(int k=1; k< h[j];k++) {
					num[k]++;
				}
			}
			for(int k=1; k<=i; k++) {
			if(max < num[k] * k) max = num[k] * k;
		}
	}

	printf("%d\n",max);

}
return 0;
}