ZOJ2347 Squares （POJ2002 二分查找）

给 N 个点，任意取其中4个，是否能构成正方形，能的话算一组，求有多少组这样的点，也就是说最多可以找出多少个正方形
方法很简单，确定两个点，必然可以求出正方形的另外两个点，在点集中查找，另外两个点是否存在就可以了。用二分查找，中间加上剪枝，时间可以压到 1s 一下，但是那些 100ms+ 的神人是怎么做到的，就不得而知了……无限Orz……
PS: ZOJ 和 POJ 输入输出格式不太一样，注意啊，好坑爹的……

递归版二分查找，最慢的一种了， ZOJ 跑了4800ms+ 才过，POJ 直接 TL ，掩面啊，泪奔啊……

#include<iostream>#include<stdio.h>#include<stdlib.h>using namespace std;#define N 1005struct point {double x,y;};point p[N];int n;bool operator ==(point a,point b)//运算符重载{if( a.x==b.x && a.y==b.y )return true;return false;}bool operator >(point a,point b){if(a.x>b.x || a.x==b.x && a.y>b.y)return true;return false;}bool operator <(point a,point b){if(a.x<b.x || a.x==b.x && a.y<b.y)return true;return false;}int cmp(const void *a,const void *b){return *(point *)a > *(point *)b ? 1:-1;}int find(point p[],int low,int high,point v)//递归的二分{if(low==high)return -1;if(p[low] == v)return low;if(p[high] == v)return high;int mid=(low+high)/2;if( p[mid] == v ) return mid;if( v > p[mid] ) return find(p,mid+1,high,v);return find(p,low,mid,v);}point Whirl(double cosl, double sinl, point a, point b)// ab 绕 a 逆时针转过角度 A 得到的点,sinl=sinA,cosl=cosA {    b.x -= a.x; b.y -= a.y;    point c;    c.x = b.x * cosl - b.y * sinl + a.x;    c.y = b.x * sinl + b.y * cosl + a.y;    return c;}int judge(int i,int j){point pa,pb,po;if(i==j)return 0;po.x=(p[i].x+p[j].x)/2.0;po.y=(p[i].y+p[j].y)/2.0;pa=Whirl(0.0,1.0,po,p[i]);//逆时针90度旋转pb=Whirl(0.0,1.0,po,p[j]);//顺时针旋转if(find(p,0,n,pa)==-1) return 0;else if(find(p,0,n,pb)==-1) return 0;return 1;}int main(){int t,i,j;scanf("%d",&t);while(t--){while(scanf("%d",&n),n){for(i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);qsort(p,n,sizeof(p[0]),cmp);int cnt=0;for(i=0;i<n;i++){for(j=i+1;j<n;j++)if(judge(i,j)==1)cnt++;}printf("%d\n",cnt/2);}if(t)puts("");}return 0;}

后来看人家的二分查找都用 while() 整，换了下，快好多，ZOJ 是 3400ms+ ，POJ 跑了 2900ms+

#include<iostream>#include<stdio.h>#include<stdlib.h>using namespace std;#define N 1005struct point {double x,y;};point p[N];int n;bool operator ==(point a,point b)//运算符重载{if( a.x==b.x && a.y==b.y )return true;return false;}bool operator >(point a,point b){if(a.x>b.x || a.x==b.x && a.y>b.y)return true;return false;}bool operator <(point a,point b){if(a.x<b.x || a.x==b.x && a.y<b.y)return true;return false;}int cmp(const void *a,const void *b){return *(point *)a > *(point *)b ? 1:-1;}bool find(point a,int m){int begin = 0,end = m-1;while( begin <= end ){int mid = ( begin + end )/2;if( a.x==p[mid].x && a.y==p[mid].y )return true;if( a.x>p[mid].x || a.x==p[mid].x && a.y>p[mid].y )begin = mid + 1;else end = mid - 1;}return false;}point Whirl(double cosl, double sinl, point a, point b)// ab 绕 a 逆时针转过角度 A 得到的点,sinl=sinA,cosl=cosA {    b.x -= a.x; b.y -= a.y;    point c;    c.x = b.x * cosl - b.y * sinl + a.x;    c.y = b.x * sinl + b.y * cosl + a.y;    return c;}int judge(int i,int j){point pa,pb,po;if(i==j)return 0;po.x=(p[i].x+p[j].x)/2.0;po.y=(p[i].y+p[j].y)/2.0;pa=Whirl(0.0,1.0,po,p[i]);//逆时针90度旋转pb=Whirl(0.0,1.0,po,p[j]);//顺时针旋转if(!find(pa,n)) return 0;else if(!find(pb,n)) return 0;return 1;}int main(){int i,j;while(scanf("%d",&n),n){for(i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);qsort(p,n,sizeof(p[0]),cmp);int cnt=0;for(i=0;i<n;i++){for(j=i+1;j<n;j++)if(judge(i,j)==1)cnt++;}printf("%d\n",cnt/2);}return 0;}

也有人用 STL 里面的二分，也挺快的，然后人又加了个剪枝，正方形只算一遍，直接压到 1000ms 左右，贴下学习下

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;struct node{int x, y;} p[1001];bool op(node xx,node yy){if (xx.x == yy.x)return xx.y < yy.y;else return xx.x < yy.x;}int main(){int points;while (scanf("%d", &points), points){int sum = 0;for (int k = 0; k < points; ++k)scanf("%d%d", &p[k].x, &p[k].y);sort(p, p + points, op);for (int i = 0; i < points; ++i){for (int j = i + 1; j < points; ++j){if(p[i].x <= p[j].x && p[i].y >= p[j].y){node p0, p1;p0.x = p[i].x + p[j].y - p[i].y;p0.y = p[i].y + p[i].x - p[j].x;p1.x = p[j].x + p[j].y - p[i].y;p1.y = p[j].y + p[i].x - p[j].x;if (!binary_search(p, p+points, p0, op))continue;if (!binary_search(p, p+points, p1, op))continue;sum++;}}}printf("%d\n", sum);}return 0;}