首页 > ACM题库 > HDU-杭电 > HDU 2830-Matrix Swapping II-动态规划-[解题报告]HOJ
2014
02-17

HDU 2830-Matrix Swapping II-动态规划-[解题报告]HOJ

Matrix Swapping II

问题描述 :

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.

输入:

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

输出:

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

样例输入:

3 4
1011
1001
0001
3 4
1010
1001
0001

样例输出:

4
2

Note: Huge Input, scanf() is recommended.

  给定一个N*M的01矩阵,在可以交换列的情况下,求出最大的全1的子矩阵。

  代码如下:

#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#define MAXN 1005
using namespace std;

int N, M, high[MAXN], temp[MAXN];

char G[MAXN][MAXN];

int deal()
{
    int Max = 0;
    memcpy(temp, high, sizeof (high));
    sort(temp+1, temp+M+1);
    for (int i = M;  i >= 1; --i) {
        if (temp[i]) {
            Max = max((M-i+1)*temp[i], Max);
        }
    }
    return Max;
}

int DP()
{
    int Max = 0;
    for (int i = 1; i <= N; ++i) {
        for (int j = 1; j <= M; ++j) {
            if (G[i][j] == '1') { 
                ++high[j];
            } 
            else {
                high[j] = 0;
            } 
        }
        Max = max(Max, deal());
    }
    return Max;
}


int main()
{
    while (scanf("%d %d", &N, &M) == 2) {
        memset(high, 0, sizeof (high));
        for (int i = 1; i <= N; ++i) {
            scanf("%s", G[i]+1);
        }
        printf("%d\n", DP());
    } 
    return 0;
}

解题参考:http://www.cnblogs.com/Lyush/archive/2012/05/01/2478118.html


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