HDU1007 求最短距离的点对

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#include<iostream>#include<cmath>#include<algorithm>#include<iomanip>using namespace std;struct sys{double x;double y;};sys num[100001],num1[100001],num2[100001];bool cmpy(sys dot1,sys dot2){if(dot1.y==dot2.y) return dot1.x<dot2.x;return dot1.y<dot2.y;}bool cmpx(sys dot1,sys dot2){if(dot1.x==dot2.x) return dot1.y<dot2.y;return dot1.x<dot2.x;}double getmin(double a,double b,double c){double minx=min(a,b);if(minx>c) minx=c;return minx;}double getdistancex(sys dot1,sys dot2){double m=sqrt((dot1.x-dot2.x)*(dot1.x-dot2.x)+(dot1.y-dot2.y)*(dot1.y-dot2.y));return m;}double getdistancey(int low,int high){int count = high -low;double dis=0;if(count==0)  return 0;else if(count==1) {dis=getdistancex(num[low],num[high]);return dis;}else if(count==2){double tem1= getdistancex(num[low],num[low+1]);double tem2= getdistancex(num[low+1],num[high]);double tem3= getdistancex(num[low],num[high]);dis=getmin(tem1,tem2,tem3); return dis;}else {double leftmin,rightmin;int mid=(high + low)/2;leftmin=getdistancey(low,mid);rightmin=getdistancey(mid+1,high);dis=min(leftmin,rightmin);int p=0;for(int i=low;i<=high;i++){if(fabs(num[i].x-num[mid].x)<dis)num1[p++]=num[i];}sort(num1,num1+p,cmpy);mid=p/2;int k=0;for(int i=0;i<p;i++){if(fabs(num1[i].y-num1[mid].y)<dis)num2[k++]=num1[i];}for(int i=0;i<k;i++){for(int j=i+1;j<k;j++){if(getdistancex(num2[i],num2[j])<dis)dis=getdistancex(num2[i],num2[j]);}}return dis;}}int main(){int n;while(cin>>n&&n!=0){for(int i=0;i<n;i++){cin>>num[i].x>>num[i].y;}sort(num,num+n,cmpx);double min=getdistancey(0,n-1);cout<<fixed<<setprecision(2)<<min/2<<endl;}return 0;}
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