二分|hash（找正方形）poj2002

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Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 16255 Accepted: 6181

Description

A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property. 

So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates. 

Input

The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.

Output

For each test case, print on a line the number of squares one can form from the given stars.

Sample Input

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

先把点从左下角到右上角排序，然后枚举两个点，利用三角形相似算出另外两个点，看是不是存在，可以用二分，也可以hash，然后答案除以2，因为我这样算，一个正方形会被算两次

二分代码：

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=1010;struct P{    int x,y;    bool operator<(const P &a)const    {        if(x==a.x)return y<a.y;        return x<a.x;    }}p[maxn];int N;int main(){    while(scanf("%d",&N)!=EOF,N)    {        for(int i=1;i<=N;i++)scanf("%d%d",&p[i].x,&p[i].y);        sort(p+1,p+1+N);        int cnt=0;        for(int i=1;i<=N;i++)        {            for(int j=i+1;j<=N;j++)            {                P tmp;                tmp.x=p[i].x-p[j].y+p[i].y;                tmp.y=p[i].y+p[j].x-p[i].x;                if(!binary_search(p+1,p+1+N,tmp))continue;                tmp.x=p[j].x-p[j].y+p[i].y;                tmp.y=p[j].y+p[j].x-p[i].x;                if(!binary_search(p+1,p+1+N,tmp))continue;                cnt++;            }        }        printf("%d\n",cnt/2);    }    return 0;}

hash：

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=1010;const int MOD=21997;int hash[MOD],next[MOD];int N;struct P{    int x,y;    bool operator<(const P &a)const    {        if(x==a.x)return y<a.y;        return x<a.x;    }}p[maxn];void insert(){    memset(hash,-1,sizeof(hash));    for(int i=1;i<=N;i++)    {        int tmp=(p[i].x*p[i].x+p[i].y*p[i].y)%MOD;        next[i]=hash[tmp];        hash[tmp]=i;    }}bool find(P tmp){    int key=(tmp.x*tmp.x+tmp.y*tmp.y)%MOD;    for(int i=hash[key];i!=-1;i=next[i])    {        if(tmp.x==p[i].x&&tmp.y==p[i].y)return true;    }    return false;}int main(){    while(scanf("%d",&N)!=EOF,N)    {        for(int i=1;i<=N;i++)scanf("%d%d",&p[i].x,&p[i].y);        sort(p+1,p+1+N);        insert();        int cnt=0;        for(int i=1;i<=N;i++)        {            for(int j=i+1;j<=N;j++)            {                P tmp;                tmp.x=p[i].x-p[j].y+p[i].y;                tmp.y=p[i].y+p[j].x-p[i].x;                if(!find(tmp))continue;                tmp.x=p[j].x-p[j].y+p[i].y;                tmp.y=p[j].y+p[j].x-p[i].x;                if(!find(tmp))continue;                cnt++;            }        }        printf("%d\n",cnt/2);    }    return 0;}