HIT2000 Maximum Submatrix（最大连续0子矩阵面积）

继续补题……

 

题目链接：

　　http://acm.hit.edu.cn/hoj/problem/view?id=2000

题目描述：

Maximum Submatrix

 

Have you ever watched the movie Matrix ? In that movie, the term Matrix does not mean a mathematical thing, but a complicated AI system.

"""

In this problem, we will go back to the original meaning of matrix. Given a 0-1 matrix, you are required to find the maximum submatrix in it which contains only 0s.

Input

An integer T (T <= 110) in the first line indicates the number of test case. For each test case, two integers N and M will be given first, then followed by a N * M 0-1 matrix.

1 <= N, M <= 100.

Output

For each test case, print the number of elements in the maximum submatrix in a single line.

Sample Input

2
2 2
0 0
0 0
4 5
1 0 1 0 0
0 1 0 0 0
0 0 1 0 0
1 1 0 0 0

Sample Output

4
8

题目大意：
　　找出最大的连续子矩阵使得其中元素均为0
思路：
　　dp
　　记录每列的前缀和，这样两数相减即可得到一列中任意连续数的和
　　二重for循环控制行的范围i~j，然后在这个范围中对于列贪心求解，类似于连续最大序列和
　　
　　复杂度O（n^3），100组数据，一秒内可做

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 110;

int main() {
    int t, m, n, ans, num[N][N], sum[N][N] = { 0 };
    cin >> t;
    while (t--) {
        ans = 0;
        scanf("%d%d", &m, &n);
        for (int i = 1; i <= m; ++i)
            for (int j = 1; j <= n; ++j) {
                scanf("%d", &num[i][j]);
                sum[i][j] = sum[i - 1][j] + num[i][j];    //记录一列中的前缀和
            }
        for (int i = 0; i < m; ++i)
            for (int j = i + 1; j <= m; ++j) {    //控制行
                int m_sum = 0, po = 1, tmp = 0;
                for (int k = 1; k <= n; ++k) {    //贪心求解
                    m_sum += sum[j][k] - sum[i][k];
                    if (m_sum)
                        tmp = 0, po = k + 1, m_sum = 0;
                    else
                        tmp += j - i, ans = max(ans, tmp);
                }
            }
        printf("%d\n", ans);
    }
}