HDU 2830 Matrix Swapping II (預處理的線性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
 

題目大意：給一個0/1矩陣，可以任意交換其中的兩列，求由1組成的最大子矩形的面積

題目分析：預處理出每個點下方有多個連續的1即cnt[i][j]，對每行的cnt值從大到小排序，枚舉列dp即可，dp[i]表示以第i行為上邊的矩形的面積最大值，轉移方程：dp[i] = max(dp[i], j * cnt[i][j])

#include 
#include 
#include 
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);
    }
}