hdu 1007(最小点对)

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分析：本题就是求解给出的点的最小点对。
代码如下：

#include <set>#include <map>#include <stack>#include <queue>#include <math.h>#include <vector>#include <string>#include <utility>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>#include <functional>using namespace std;const double pi=acos(-1);const int maxn=100005;const int INF=0x3f3f3f;struct Point{    double x,y;}p[maxn],a[maxn],b[maxn];bool cmpx(Point a,Point b){    return a.x<b.x;}bool cmpy(Point a,Point b){    return a.y<b.y;}double Cross(Point a,Point b,Point c){    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}double dot(Point a,Point b,Point c){    return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);}double dis(Point a,Point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}Point cals(Point a,Point b,double r){    Point ans;    ans.x=(a.x-b.x)*cos(r)-(a.y-b.y)*sin(r)+b.x;    ans.y=(a.x-b.x)*sin(r)+(a.y-b.y)*cos(r)+b.y;    return ans;}//旋转double solve(int s,int e){    if(s+1==e)return dis(a[s],a[e]);    if(s+2==e)return min(dis(a[s],a[s+1]),min(dis(a[s+1],a[e]),dis(a[s],a[e])));    int mid=(s+e)>>1;    double ans=min(solve(s,mid),solve(mid+1,e));    int cnt=0;    for(int i=s;i<=e;i++){        if(a[i].x>=a[mid].x-ans&&a[i].x<=a[mid].x+ans)        b[cnt++]=a[i];    }    sort(b,b+cnt,cmpy);    for(int i=0;i<cnt;i++){        for(int j=i+1;j<cnt;j++){            if(b[j].y-b[i].y>=ans)            break;            ans=min(ans,dis(b[j],b[i]));        }    }    return  ans;}int main(){    int n;    while(scanf("%d",&n)!=EOF&&n){        for(int i=0;i<n;i++){            scanf("%lf%lf",&p[i].x,&p[i].y);            a[i]=p[i];        }        sort(a,a+n,cmpx);        printf("%.2f\n",solve(0,n-1)/2);    }    return 0;}

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