BIT1033 POJ 2002 Squares

SET解法（此法在POJ会TLE，哈希解法在后面）

题目的意思是

给n个点，让在n个点中找四个点为正方形的方案数，同四个点旋转，反转什么的只能算一个

一共有1000个点

如果直接枚举是

1000×1000×1000×1000的复杂度

每次枚举两个点，再用两个点算出另两个点的坐标，看这两个点的坐标存不存在

这样的复杂度就是1000×1000×log(1000)因为我是用set 实现查找的

已知1，2点的坐标，这里我们确认1，2是相邻的两点

这样一个正方形会被计算四次

已知： (x1,y1)  (x2,y2)

则：   x3=x1+(y1-y2)   y3= y1-(x1-x2)

x4=x2+(y1-y2)   y4= y2-(x1-x2)

或

x3=x1-(y1-y2)   y3= y1+(x1-x2)

x4=x2-(y1-y2)   y4= y2+(x1-x2)

接下来上代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<set>
class coordinate
{
public:
	int x,y;
};
using namespace std;
coordinate a[1010];
bool operator<(coordinate a,coordinate b)
{
	if(a.x==b.x)
	{
		return a.y<b.y;
	}
	return a.x<b.x;
}
int main()
{
	int n;
	while (scanf("%d",&n),n)
	{
		set<coordinate>mySet;
		for (int i = 0; i < n; i++)
		{
			coordinate tt;
			scanf("%d %d",&tt.x,&tt.y);
			a[i]=tt;
			mySet.insert(tt);
		}
		int counter=0;
		for (int i = 0; i < n; i++)
		{
			for (int j = i+1; j < n; j++)
			{
				//coordinate[i] coordinate[j]
				coordinate t1,t2;
				t1.x=a[i].x+a[i].y-a[j].y;
				t1.y=a[i].y-a[i].x+a[j].x;
				t2.x=a[j].x+a[i].y-a[j].y;
				t2.y=a[j].y-a[i].x+a[j].x;
				if(mySet.count(t1)&&mySet.count(t2))
				{
					/*cout<<a[i].x<<' '<<a[i].y<<endl;
					cout<<a[j].x<<' '<<a[j].y<<endl;
					cout<<t1.x<<' '<<t1.y<<endl;
					cout<<t2.x<<' '<<t2.y<<endl;
					cout<<endl;*/
					counter++;
				}
				t1.x=a[i].x-a[i].y+a[j].y;
				t1.y=a[i].y+a[i].x-a[j].x;
				t2.x=a[j].x-a[i].y+a[j].y;
				t2.y=a[j].y+a[i].x-a[j].x;
				if(mySet.count(t1)&&mySet.count(t2))
				{
					/*cout<<a[i].x<<' '<<a[i].y<<endl;
					cout<<a[j].x<<' '<<a[j].y<<endl;
					cout<<t1.x<<' '<<t1.y<<endl;
					cout<<t2.x<<' '<<t2.y<<endl;
					cout<<endl;*/
					counter++;
				}
			}
		}
		printf("%d\n",counter/4);
	}
	return 0;
}

HASH解法：

用哈希表替换掉set，就可以在poj过了

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<list>
using namespace std;
class coordinate
{
public:
	int x,y;
};
class T
{
public:
    list<coordinate>Hash[60000];//2*x+y
    void Insert(coordinate t)
    {
        Hash[(t.x*2+t.y+60000)%60000].push_back(t);
    }
    void Clear()
    {
        for(int i=0;i<60000;i++)
        {
            Hash[i].clear();
        }
    }
    bool Count(coordinate t)
    {
        int HashNum=(t.x*2+t.y+60000)%60000;
        for(list<coordinate>::iterator i=Hash[HashNum].begin();i!=Hash[HashNum].end();i++)
        {
            if(i->x==t.x&&i->y==t.y)
            {
                return true;
            }
        }
        return false;
    }
}MySet;
coordinate a[1010];
bool operator<(coordinate a,coordinate b)
{
	if(a.x==b.x)
	{
		return a.y<b.y;
	}
	return a.x<b.x;
}
int main()
{
	int n;
	while (scanf("%d",&n),n)
	{
	    MySet.Clear();
		for (int i = 0; i < n; i++)
		{
			coordinate tt;
			scanf("%d %d",&tt.x,&tt.y);
			a[i]=tt;
			MySet.Insert(tt);
		}
		int counter=0;
		for (int i = 0; i < n; i++)
		{
			for (int j = i+1; j < n; j++)
			{
				coordinate t1,t2;
				t1.x=a[i].x+a[i].y-a[j].y;
				t1.y=a[i].y-a[i].x+a[j].x;
				t2.x=a[j].x+a[i].y-a[j].y;
				t2.y=a[j].y-a[i].x+a[j].x;
				if(MySet.Count(t1)&&MySet.Count(t2))
				{
					counter++;
				}
				t1.x=a[i].x-a[i].y+a[j].y;
				t1.y=a[i].y+a[i].x-a[j].x;
				t2.x=a[j].x-a[i].y+a[j].y;
				t2.y=a[j].y+a[i].x-a[j].x;
				if(MySet.Count(t1)&&MySet.Count(t2))
				{
					counter++;
				}
			}
		}
		printf("%d\n",counter/4);
	}
	return 0;
}