topcoder-srm610-div2-550（暴力乱搞）

给你一个01矩阵，求其中最大的01交替的矩阵
由于n最大才100，所以直接暴力乱搞
先求出第i行，所有列往上的合法长度，然后枚举以第j列为最左边的列，计算可以得到的最大矩阵

/*************************************************************************
    > File Name: 2.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月07日 星期四 15时07分58秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int ok[110][110];

class TheMatrix {
    public:
        int MaxArea(vector <string> mat) {
            memset(ok, 0, sizeof(ok));
            int n = mat.size();
            int m = mat[0].length();
            for (int i = 0; i < m; ++i) {
                ok[i][0] = 1;
                for (int j = 1; j < n; ++j) {
                    if (mat[j][i] == mat[j - 1][i]) {
                        ok[i][j] = 1;
                    }
                    else {
                        ok[i][j] = ok[i][j - 1] + 1; //the ith row of jth col
                    }
                }
            }
            int size = 0;
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    int mins = ok[j][i];
                    size = max(size, mins);
                    int flag = mat[i][j] - ‘0‘;
                    for (int k = j + 1; k < m; ++k) {
                        if (flag == mat[i][k] - ‘0‘) {
                            break;
                        }
                        flag ^= 1;
                        mins = min(mins, ok[k][i]);
                        size = max(size, (k - j + 1) * mins);
                    }
                }
            }
            return size;
        }
};