正方形（squares,UVa201)

这个题目思路还是比较简单的，两个数组储存行和列的情况。然后再遍历各个顶点的情况就可以了。其他的见源码中的注释。

#include
#include
using namespace std;
struct zuobao              //创建结构体用来储存横边和竖边的坐标
{
    int row;
    int column;
};
bool ifexist(int i, int j, int size, const zuobao* rc)
{
    for (int m = 0; m < size; ++m)
    {
        if (rc[m].row == i && rc[m].column == j)
            return true;
    }
    return false;
}
void panduan(int i, int j, int square_size, int lenth, int *number, const zuobao *r, const zuobao*c)    // i为横坐标，j为纵坐标，该函数用来计算以该点为右上角的起点，可以构成什么正方形
{

    for (int m = 1; m < square_size; ++m)           //循环判断规格为1--n-1的正方形
    {
        int key = 1;                                    //用来标记边是否存在
        for (int n = 0; n < m; ++n)                          //左边竖向的边
        {
            if (!ifexist(i + n, j, lenth, c))
            {
                key = 0;
                break;                             //缺少边跳出循环
            }
        }
        if (key == 0)                             //如果缺少跳过下面的判断
            continue;
        for (int n = 0; n < m; ++n)
        {
            if (!ifexist(i + m, j + n, lenth, r))
            {
                key = 0;
                break;
            }
        }
        if (key == 0)
            continue;
        for (int n = 0; n < m; ++n)
        {
            if (!ifexist(i + n, j + m, lenth, c))
            {
                key = 0;
                break;
            }
        }
        if (key == 0)
            continue;
        for (int n = 0; n < m; ++n)
        {
            if (!ifexist(i, j + n, lenth, r))
            {
                key = 0;
                break;
            }
        }
        if (key == 0)
            continue;
        ++number[m];
    }
}
int main()
{
    ///*文件重定向，输出测试*/
    //ifstream fin;
    //fin.open("data.in");
    //cin.rdbuf(fin.rdbuf());
    //ofstream out;
    //out.open("data.out");
    //cout.rdbuf(out.rdbuf());
    int n = 0, m = 0, times = 1;       //n为正方体边上点的个数，m为边的个数
    int t = 0;
    while (cin >> n >> m)
    {
        if (0 != t)
            cout <> kind;
            if(kind=='H')
                cin >> r[i].row >> r[i].column;
            else
                cin >> c[i].column >> c[i].row;
        }                                             //注意这里题目里的i变成了列，而j变成了行，注意调换顺序
        cout << "Problem #" << times << endl << endl;
        ++times;
        for (int i = 1; i < n; ++i)
            for (int j = 1; j < n; ++j)
            {
                panduan(i, j, n, m, number, r, c);
            }
        int temp = 0;
        for (int i = 1; i < n; ++i)
        {
            if (number[i] != 0)
            {
                cout << number[i] << " square (s) of size " << i << endl;
                temp = 1;
            }
        }
        if (0 == temp)
            cout << "No completed squares can be found." << endl;
    

    }
}