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;}