【hdu1007】最近点对

http://acm.hdu.edu.cn/showproblem.php?pid=1007

分治法的经典应用，复杂度可以证明为nlognlogn

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <map>
 #include <stack>
 #include <string>
 #include <ctime>
 #include <queue>
 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem(a, b) memset(a, b, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define eps 0.0000001
 #define lowbit(x) ((x) & -(x))
 #define memc(a, b) memcpy(a, b, sizeof(b))
 #define x_x(a) ((a) * (a))
 using namespace std;
 #define LL __int64
 #define DB double
 #define pi 3.14159265359
 #define MD 10000007
 #define INF (int)1e9
 struct Point {
         double x, y;
         Point(double x = , double y = ):x(x), y(y) {}
         void inp() {
                 scanf("%lf%lf", &x, &y);
         }
         bool operator < (const Point &a) const {
                 return x < a.x || x == a.x && y < a.y;
         }
 } pn[], tmp[];
 int cmpy(Point i, Point j)
 {
         return i.y < j.y;
 }
 double dist(Point a, Point b)
 {
         return sqrt(x_x(a.x - b.x) + x_x(a.y - b.y));
 }
 double solve(int l, int r)
 {
         if(l == r) return INF;
         int m = (l + r) >> ;
         double d1 = solve(l, m), d2 = solve(m + , r), d = min(d1, d2);
         int nn = ;
         for(int i = l; i <= r; i++) {
                 if(dist(pn[i], pn[m]) <= d) {
                         tmp[++nn] = pn[i];
                 }
         }
         sort(tmp + , tmp + nn, cmpy);
         for(int i = ; i < nn; i++) {
                 double mind = INF;
                 for(int j = i + ; j <= nn && tmp[j].y - tmp[i].y <= d; j++) {
                         mind = min(mind, dist(tmp[i], tmp[j]));
                 }
                 d = min(d, mind);
         }
         return d;
 }
 int main()
 {
         //freopen("input.txt", "r", stdin);
         //freopen("output.txt", "w", stdout);
         int n;
         while(cin>> n, n) {
                 for(int i = ; i <= n; i++) pn[i].inp();
                 sort(pn + , pn +  + n);
                 double t = solve(, n);
                 printf("%.2f\n", t / );
         }
         return ;
 }