poj- 2002-Squares-哈希|除法散列法

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题意：

给你n 个点的坐标，让你去算一下，能够形成多少个正方形

思路：

1.枚举两个点，然后推出那两个的坐标，这样会有重复的，最后需要除以4

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)

数学公式，也是看的网上的，自己没去推，同学们不要学我啊

2. 点的寻找要用哈希的方法，链地址法还是啥搞不懂，我用的是数组模拟链表，除法散列法

求 key 的值的时候用的平方求于法

#include<stdio.h>#include<string.h>#include<stdlib.h>#define mod 100007int tmp;struct node{    int x, y;    int next;}ls[1010],a[1010];int head[mod];void add(int x,int y){    int key = (x*x + y*y)%mod;    if(head[key]==-1)    {        ls[tmp].x = x;        ls[tmp].y = y;        ls[tmp].next = -1;        head[key] = tmp++;    }    else    {        int c = head[key];        while(1)        {            if(ls[c].x == x && ls[c].y == y)            {                return;            }            if(ls[c].next == -1)                break;            c = ls[c].next;        }        ls[tmp].x = x;        ls[tmp].y = y;        ls[tmp].next = -1;        ls[c].next = tmp++;    }}int sear(int x, int y){    int ad = (x*x +y*y)%mod;    int c= head[ad];    while(c!=-1)    {        if(ls[c].x == x && ls[c].y == y)            return 1;        c = ls[c].next;    }    return 0;}int main(){    int n;    while(~scanf("%d",&n)&&n)    {        tmp = 1;        memset(head,-1,sizeof(head));        int i, j;        for(i = 0; i < n; i++)        {            int x, y;            scanf("%d%d",&x,&y);            a[i].x = x;            a[i].y = y;            add(x,y);        }        int x3, y3, x4, y4;        int cnt = 0;        for(i = 0; i < n-1; i++)        {            for(j = i+1; j < n; j++)            {               x3 = a[i].x - (a[i].y-a[j].y);               y3 = a[i].y + (a[i].x-a[j].x);               x4 = a[j].x - (a[i].y - a[j].y);               y4 = a[j].y + (a[i].x -a[j].x);               if(sear(x3,y3)&&sear(x4,y4))                cnt++;               x3 = a[i].x + (a[i].y-a[j].y);               y3 = a[i].y - (a[i].x-a[j].x);               x4 = a[j].x + (a[i].y - a[j].y);               y4 = a[j].y - (a[i].x -a[j].x);               if(sear(x3,y3)&&sear(x4,y4))                cnt++;            }        }        printf("%d\n",cnt/4);    }    return 0;}

map

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<queue>#include<climits>#include<list>#define MULT 20000using  namespace std;struct node{    int x,y;}st[1200];int main(){    int n;    while(scanf("%d",&n),n!=0)    {        map<pair<int ,int>,int>hash_list;        int i;        int j;        for(i=0;i<n;++i)        {            scanf("%d%d",&st[i].x,&st[i].y);            hash_list[pair<int,int>(st[i].x,st[i].y)];        }        int sum=0;        for(i=0;i<n;++i)        {            for(j=i+1;j<n;++j)            {                int  x,y,xx,yy;                x=(st[i].x+st[j].x+st[i].y-st[j].y);                y=(st[i].y+st[j].y+st[j].x-st[i].x);                xx=(st[i].x+st[j].x-st[i].y+st[j].y);                yy=(st[i].y+st[j].y-st[j].x+st[i].x);                if(x&1||y&1||xx&1||yy&1)                {                    continue;                }                x/=2;                y/=2;                xx/=2;                yy/=2;                if(hash_list.count(pair<int,int>(x,y))&&hash_list.count(pair<int,int>(xx,yy)))                {                    sum++;                }            }        }        cout<<sum/2<<endl;    }}

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