UVa 201 正方形 算法竞赛入门经典 习题4-2

注释是用于调试的，请忽视。规模这么小，暴力就完事了。本题大可不必用结构体，一开始用结构体是想转化成连接矩阵做，然后找规律找了半天没找到。。。。。。比较简单，易错点就是v（ i ，j） 代表的意思是 第i列第j行的坐标为起点。

dev c++能不能一键复制所有输出啊。。。。我只会把答案输出到文件复制，求解！

uva201题目原址
该题样例调试

#include<iostream>
#include<string.h>
//#include<fstream>
using namespace std;
int aa[105][105];
struct bz
{
	char c;
	int x;
	int y;
}e[205];
int main()
{
	//ofstream fout;
//	fout.open("D:\\anwser.txt");
	int t = 0;
	int n = 0;
	int b = 0;
	while(cin>>n && cin>>b)
	{
		for(int i = 0;i<105;++i)
		   memset(aa[i],0,sizeof(aa[i]));
		for(int i = 0;i<b;++i)
		 {
		 	cin>>e[i].c;
		 	cin>>e[i].x>>e[i].y;
		 	if(e[i].c == 'H') aa[e[i].x][e[i].y] +=1;
		 	else aa[e[i].y][e[i].x] +=2;
		 }
		 int a[10] = {0};
		 for(int i = 1;i<=n;++i)
		 {
		   for(int j = 1;j<=n;++j)
		   {
		   	for(int k = 1;k<=n-i;++k)
		   	{
		   		int ff = 1;
		   		for(int p = i;p<=i+k-1 && ff == 1;++p) 
		   		  {
		   		  	if(aa[p][j] == 1 || aa[p][j] == 0)
		   		  	   ff = 0;
		   		  	if(aa[p][j+k] == 1 || aa[p][j+k] == 0)
		   		  	   ff = 0;
					 }
			    for(int p = j;p<=j+k-1 && ff == 1;++p)
			     {
			     	if(aa[i][p] == 2 || aa[i][p] == 0)
			     	    ff = 0;
			     	if(aa[i+k][p] == 2 || aa[i+k][p] == 0)
			     	    ff = 0;
				 }
				 if(ff) 
				 ++a[k];
			}
		   }
		  }
		  if(t) {
		  	cout<<endl;
		   // fout<<endl;
		  	cout<<"**********************************"<<endl;
		   // fout<<"**********************************"<<endl;
		  	cout<<endl;
		  	//fout<<endl;
		  
		  }
		  cout<<"Problem #"<<++t<<endl;
		  //fout<<"Problem #"<<t<<endl;
		  cout<<endl;
		  //fout<<endl;
		  int y = 0;
		   for(int i = 1;i<10;++i)
		   {
		   	 if(a[i]) {
		   	 	 cout<<a[i]<<' '<<"square (s) of size"<<' '<<i<<endl;
		   	 	 //fout<<a[i]<<' '<<"square (s) of size"<<' '<<i<<endl;
		   	 	 y = 1;
				}
		   }
		   if(!y) {
		   	cout<<"No completed squares can be found."<<endl;
		   	//fout<<"No completed squares can be found."<<endl;
		   }
	}
//	fout.close();
}