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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
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题目大意：给一个0/1矩阵，可以任意交换其中的两列，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即可，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
#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); } }
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