HDU 2830 Matrix Swapping II

HDU 2830 Matrix Swapping II

Time Limit: 9000/3000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)

Problem Description

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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

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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

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Output one line for each test case, indicating the maximum possible goodness.

Sample Input

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3 4
1011
1001
0001
3 4
1010
1001
0001

Sample Output

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4
2

Note

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Huge Input, scanf() is recommended.

Source

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2009 Multi-University Training Contest 2 - Host by TJU

Recommend

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gaojie

Submit

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define MAXN 1005bool cmp(int a, int b){    return a > b;}int main()//这是一道DP题{    int N, M;    int i, j;    int visit[MAXN], num[MAXN];//visit代表到某点列上连续的1的数量    char str[MAXN];    int ans;    while(~scanf("%d %d", &N, &M))    {        ans = 0;        memset(visit, 0, sizeof(visit));        memset(num, 0, sizeof(num));        for(i = 0; i < N; i++)        {            scanf("%s", str);            for(j = 0; j < M; j++)            {                int t = str[j] - '0';                if(t)//如果是1的话，在第i行第j列就多了一个连续的1                    visit[j]++;                else//否则在第i行第j列没有连续的1                    visit[j] = 0;                num[j] = visit[j];//用num数组单独存储这一行的visit            }            sort(num, num+M, cmp);//因为列是可交换的，所有可以把num从大到小排序，可以在x-y坐标系中想象num数组            for(j = 0; j < M; j++)                ans = max(ans, num[j]*(j+1));//矩形面积S=y*x，y是num[j],x是(j+1),找到最大的矩形面积，每输入1行就计算1次ans        }        printf("%d\n", ans);    }    return 0;}