topcoder-srm610-div2-550（暴力乱搞）

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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;        }};

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