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

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本文介绍了一道算法题，给定一个0/1矩阵，可通过任意交换两列来最大化由1组成的最大子矩形面积。文章提供了详细的解题思路与代码实现。

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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\n", ans);
    }
}