HDU 2830 Matrix Swapping II (预处理的线性dp)

Matrix Swapping II

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

Total Submission(s): 1430    Accepted Submission(s): 950

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.

 

Source

2009 Multi-University Training Contest 2 - Host by TJU

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2830

题目大意：给一个0/1矩阵，能够随意交换当中的两列，求由1组成的最大子矩形的面积

题目分析：预处理出每一个点下方有多个连续的1即cnt[i][j]。对每行的cnt值从大到小排序。枚举列dp就可以，dp[i]表示以第i行为上边的矩形的面积最大值。转移方程：dp[i] = max(dp[i], j * cnt[i][j])

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 1e3 + 5;
int const INF = 0x3fffffff;
char s[MAX][MAX];
int cnt[MAX][MAX];
int dp[MAX];
int n, m;

bool cmp(int a, int b)
{
    return a > b;
}

int main()
{
    while(scanf("%d %d", &n ,&m) != EOF)
    {
        memset(cnt, 0, sizeof(cnt));
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
            scanf("%s", s[i] + 1);
        for(int i = n; i >= 1; i--)
            for(int j = 1; j <= m; j++)
                if(s[i][j] - '0')
                    cnt[i][j] = cnt[i + 1][j] + 1;
        for(int i = 1; i <= n; i++)
        {
            sort(cnt[i] + 1, cnt[i] + 1 + m, cmp);
            for(int j = 1; j <= m; j++)
                if(cnt[i][j])
                    dp[i] = max(dp[i], j * cnt[i][j]);
        }
        int ans = 0;
        for(int i = 1; i <= n; i++)
            ans = max(ans, dp[i]);
        printf("%d
", ans);
    }
}