hdu 1007_平面最近点对模板

简单题目,直接模板上

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

O(n * log(n) *log(n) )的复杂度

#include <cstdio>#include <ctime>#include <cstdlib>#include <cstring>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <vector>#include <iostream>#include <algorithm>#include <bitset>#include <fstream>using namespace std;//LOOP#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FED(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)//OTHER#define SZ(V) (int)V.size()#define PB push_back#define MP make_pair#define all(x) (x).begin(),(x).end()//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define RS(s) scanf("%s", s)//OUTPUT#define WI(n) printf("%d\n", n)#define WS(n) printf("%s\n", n)//debug//#define online_judge#ifndef online_judge#define debugt(a) cout << (#a) << "=" << a << " ";#define debugI(a) debugt(a) cout << endl#define debugII(a, b) debugt(a) debugt(b) cout << endl#define debugIII(a, b, c) debugt(a) debugt(b) debugt(c) cout << endl#define debugIV(a, b, c, d) debugt(a) debugt(b) debugt(c) debugt(d) cout << endl#else#define debugI(v)#define debugII(a, b)#define debugIII(a, b, c)#define debugIV(a, b, c, d)#endif#define sqr(x) (x) * (x)typedef long long LL;typedef unsigned long long ULL;typedef vector <int> VI;const int INF = 0x3f3f3f3f;const double EPS = 1e-10;const int MOD = 100000007;const int MAXN = 100010;const double PI = acos(-1.0);inline int dcmp(double x){    if(fabs(x) < EPS) return 0;    else return x < 0 ? -1 : 1;}struct Point{    double x, y;    Point(double x=0, double y=0):x(x),y(y) { }    inline void read()    {        scanf("%lf%lf", &x, &y);    }};//最近点对Point point[MAXN];int tmpt[MAXN], Y[MAXN];inline bool cmpxy(Point a, Point b){    if(a.x != b.x)        return a.x < b.x;    return a.y < b.y;}inline bool cmpy(int a, int b){    return point[a].y < point[b].y;}inline double dist(int x, int y){    Point& a = point[x], &b = point[y];    return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}double Closest_Pair(int left, int right){    double d = 1e100;    if(left == right)        return d;    if(left + 1 == right)        return dist(left, right);    int mid = (left + right) >> 1;    double d1 = Closest_Pair(left, mid);    double d2 = Closest_Pair(mid + 1, right);    d = min(d1, d2);    int k = 0;    //分离出宽度为d的区间    FE(i, left, right)    {        if(fabs(point[mid].x - point[i].x) <= d)            tmpt[k++] = i;    }    sort(tmpt, tmpt + k, cmpy);    //线性扫描    REP(i, k)    {        for(int j = i + 1; j < k && point[tmpt[j]].y-point[tmpt[i]].y < d; j++)        {            double d3 = dist(tmpt[i],tmpt[j]);            if(d > d3)                d = d3;        }    }    return d;}int main(){    //freopen("input.txt", "r", stdin);    int n;    while (~RI(n) && n)    {        REP(i, n)        {            point[i].read();        }        sort(point, point + n, cmpxy);        printf("%.2f\n", Closest_Pair(0, n - 1) / 2);    }    return 0;}