HDU 1007 Quoit Design

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即扔圈套物，求所有点中距离最近的两点，这两点距离为环的直径，一般用二分法。下面是比较神奇的方法，不是二分

与二分相同的是都要按x和y排序

解法二与二分思路：http://blog.csdn.net/hellobabygogo3/article/details/8042650

#include<stdio.h>#include<stdlib.h>#include<math.h>struct node{double x,y;}a[100005];int cmpxy( const void *a , const void *b ){struct node *c = (struct node *)a;struct node *d = (struct node *)b;if( fabs(c->x-d->x)>1e-8 ) return c->x > d->x ? 1 : -1;//比较x相同用fabs()else return c->y > d->y ? 1 : -1;}double min2(double x,double y){return x>y?y:x;}double dis(struct node a,struct node b)//参数类型坑死人，struct不能省{return sqrt( (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) );}double f( struct node a[],int n){int i;double d = 1e+10,td;if(n == 2) return dis(a[0],a[1]);for(i=0;i<n-2;i++){td = min2( dis(a[i],a[i+1]) , dis(a[i],a[i+2]) );td = min2( td , dis(a[i+1],a[i+2]) );d = min2( td,d);}return d;}int main(){int i,n;//freopen("a.txt","r",stdin);while( scanf("%d",&n)!=EOF && n ){for(i=0;i<n;i++)scanf("%lf%lf",&a[i].x,&a[i].y);qsort(a,n,sizeof(a[0]),cmpxy);printf("%.2lf\n",f(a,n)/2);}return 0;}

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