poj 2002 Squares【HASH】POJ最快

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先排序，然后枚举任意两点(x1,y1)(x2,y2)，则如果存在点(x1+y1-y2,y1-x1+x2)(x2+y1-y2,y2-x1+x2)则它们能构成一个正方形

157MS

#include<cstring>#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int maxn=10000+10;struct node{    int a,b,c;}b[10000];#define hash(x) ((x)&131071)int a[141000];void init(){    memset(a,255,sizeof(a));}int ins(int x){    int z=hash(x);    while(a[z]!=-1&&a[z]!=x) z=hash(z+1);    if(a[z]==-1) a[z]=x;    return z;}int get(int x){    int z=hash(x);    while(a[z]!=-1&&a[z]!=x) z=hash(z+1);    if(a[z]==-1) return -1;    return z;}void read(int &d){    char ch;bool flag=0;    while(ch=getchar(),(ch>57||ch<48)&&ch!=45);    ch==45?flag=1,d=0:d=ch-48;    while(ch=getchar(),ch<=57&&ch>=48) d=d*10+ch-48;    if(flag) d=-d;}bool cmp(node a,node b){    return a.a==b.a?a.b<b.b:a.a<b.a;}int main(){    int n;    while(read(n),n)    {        int ans=0;        init();        for(int i=0;i<n;i++)        {            read(b[i].a);read(b[i].b);            b[i].a+=20500,b[i].b+=20500;            b[i].c=b[i].a*41000+b[i].b;            ins(b[i].c);        }        int x3,y3,x4,y4;        sort(b,b+n,cmp);        for(int i=0;i<n;i++)        {            int &x1=b[i].a,&y1=b[i].b;            for(int j=i+1;j<n;j++)            {                int &x2=b[j].a,&y2=b[j].b;                x3=x1+y1-y2,y3=y1+x2-x1;                x4=x2+y1-y2,y4=y2+x2-x1;                if(get(x3*41000+y3)==-1||get(x4*41000+y4)==-1) continue;                ans++;            }        }        printf("%d\n",ans/2);    }    return 0;}

((x<<9)^y)&p p=65535或131071 157MS

#include<cstdio>#include<iostream>#include<algorithm>using namespace std;#define hash(x) ((x)&131071)int a[140000][2];pair<int,int>b[10000];int main(){    int n,ph;    int x,y,dx,dy;    while(scanf("%d",&n)&&n)    {        int ans=0;        memset(a,255,sizeof(a));        for(int i=0;i<n;i++)        {            scanf("%d%d",&x,&y);            x+=30000;y+=30000;            b[i].first=x,b[i].second=y;            ph=hash((x<<9)^y);            while(a[ph][0]!=-1) ph=hash(ph+1);            a[ph][0]=x,a[ph][1]=y;        }        sort(b,b+n);        for(int i=0;i<n;i++)            for(int j=i+1;j<n;j++)            {                dx=b[j].first-b[i].first;                dy=b[i].second-b[j].second;                x=b[i].first+dy;                y=b[i].second+dx;                ph=hash((x<<9)^y);                while(a[ph][0]!=-1&&!(a[ph][0]==x&&a[ph][1]==y)) ph=hash(ph+1);                if(a[ph][0]==-1) continue;                x=b[j].first+dy;                y=b[j].second+dx;                ph=hash((x<<9)^y);                while(a[ph][0]!=-1&&!(a[ph][0]==x&&a[ph][1]==y)) ph=hash(ph+1);                if(a[ph][0]==-1) continue;                ans++;            }        printf("%d\n",ans/2);    }    return 0;}