分治——最小点对（HDU 1007）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1007
题意：给出一个二维平面和N个点，求两点间的最小距离
分析：分治即可
P.S.：发现一个大坑！！！！！#define MIN(x,y) x < y?x:y 只能传变量，不能传函数和其他表达式！！！否则比较的时候计算一次，出来又计算一次，绝壁超时啊！！！！！在这TLE了3个小时刷了2页submition啊啊啊！
AC代码

/*************************************************************************    > File Name: 1007.cpp    > Author: Akira    > Mail: qaq.febr2.qaq@gmail.com    > Created Time: 2017年01月17日 星期二 10时44分03秒 ************************************************************************/#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cstdlib>#include<algorithm>#include<queue>#include<stack>#include<map>#include<cmath>#include<vector>#include<set>#include<list>#include<ctime>typedef long long LL;typedef unsigned long long ULL;typedef long double LD;#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define Sqr(a) ((a)*(a))using namespace std;#define MaxN 100001#define MaxM MaxN*10#define INF 0x3f3f3f3f#define EPS 1e-8#define bug cout << 88888888 << endl;#define MIN(x,y) (x<y?x:y)#define MAX(x,y) (x>y?x:y)int N;struct Point{double x,y;}point[MaxN];bool cmp (Point a, Point b){    if(a.x == b.x) return a.y<b.y;    return a.x<b.x;}bool cmpQ(Point a, Point b){    if(a.y==b.y) return a.x<b.x;    return a.y>b.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) );}double check(double d, int mid,int l ,int r){    vector<Point> V;    double midx = point[mid].x;    for(int i=l;i<=r;i++)    {        if( fabs(point[i].x - midx) <d  ) V.push_back(point[i]);    }    mid = V.size()/2;    vector<Point> CV;    for(int i=0;i<V.size();i++)    {        if(fabs(V[i].y - V[mid].y) <d) CV.push_back(V[i]);    }    sort(CV.begin(), CV.end(), cmpQ);    double ans = d;    for(int i=0;i<CV.size();i++)    {        for(int j=i+1;j<CV.size()&&j<=i+5;j++)        {            ans = min(ans, dis(CV[i], CV[j]));        }    }    return ans;}double solve(int l, int r){    if(r-l>2)    {        int mid = (r+l)/2;        double ans = min(solve(l,mid), solve(mid+1,r));        ans = check(ans, mid,l,r);        return ans;    }    else    {        double ans = INF*1.0;        for(int i=l;i<=r;i++)        {            for(int j=i+1;j<=r;j++)            {                ans = MIN(ans, dis(point[i],point[j]));            }        }        return ans;    }}int main(){    int N;    while(~scanf("%d", &N) && N)    {        for(int i=1;i<=N;i++)        {            scanf("%lf%lf", &point[i].x, &point[i].y);        }        sort(point+1, point+1+N, cmp);        double ans = solve(1,N);        printf("%.2lf\n", ans/2.0);    }}

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