HDU 2830 Matrix Swapping II (linear dp of preprocessing)

HDU 2830 Matrix Swapping II (linear dp of preprocessing)

 

Matrix Swapping II Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission (s): 1430 Accepted Submission (s): 950Problem 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 one line for each test case, indicating the maximum possible goodness.
Sample Input

3 41011100100013 4101010010001

Sample Output

42Note: Huge Input, scanf() is recommended.

Source 2009 Multi-University Training Contest 2-Host by TJU
 

Given a 0/1 matrix, two columns can be exchanged at will to obtain the area of the largest subrectangle consisting of 1.

Question Analysis: pre-process multiple consecutive 1 (cnt [I] [j]) under each vertex. Sort the cnt values of each row in ascending order and enumerate the column dp, dp [I] indicates the maximum area of the rectangle above the I behavior. The transfer equation is 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);    }}