uva 201 数正方形数

http://acm.bnu.edu.cn/v3/contest_show.php?cid=5772#problem/B

给你一个由点组成的网格，再给一些连结相邻点的操作（横向、纵向），统计里面的正方形，从小到大输出其数量。

水题，依次枚举边长，判断每个点为左上角时能否构成正方形..

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <sstream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define clr1(x) memset(x,-1,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
const int inf = 1000000000;
const int maxn = 15;

int h[maxn][maxn],v[maxn][maxn],n,m;
int main()
{
    int _ = 0;
    while(~RD2(n,m)){
        int x,y;
        char typ[2];
        clr0(h),clr0(v);
        while(m--){
            scanf("%s",typ);
            RD2(x,y);
            if(typ[0] == 'H')
                h[x][y] = 1;
            else
                v[y][x] = 1;
        }
        if(_++) puts("\n**********************************\n");
        printf("Problem #%d\n\n",_);

        int sum = 0;
        for(int l = 1;l <= n;++l){
            int cnt = 0;
            for(int i = 1;i + l <= n;++i)
            for(int j = 1;j + l <= n;++j){
                int flag = 1;
                for(int hh = j;hh < j + l;++hh)
                    if(!h[i][hh] || !h[i+l][hh])
                        flag = 0;
                for(int vv = i;vv < i + l;++vv)
                    if(!v[vv][j] || !v[vv][j+l])
                        flag = 0;
                cnt += flag;
            }
            sum += cnt;
            if(cnt)
                printf("%d square (s) of size %d\n",cnt,l);
        }
        if(!sum)
            puts("No completed squares can be found.");
    }
    return 0;
}