uva 201 数正方形数

来源：互联网发布：linux查看进程占用内存编辑：程序博客网时间：2024/04/27 14:28

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-9const 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;}

0 0