HDOJ 1007 Quoit Design

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题意：有若干组数据，每组数据第一行为n，后面有n行，每行有两个数据，代表一个点的坐标x,y。求这些点之间最近两点的距离的一半，结果保留两位小数。

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

思路：分治，将所有点分成两部分，有些特殊情况，一点在左区域，另一点在右区域，在已知的最短距离一半的范围内查找是否存在更短的点对（参考：http://www.cnblogs.com/AdaByron/archive/2011/10/07/2200966.html）。

注意点：cin关闭流同步也会超时。用scanf能过。

以下为AC代码：

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor129281222015-02-10 15:08:47Accepted10071216MS4340K2265 BG++luminous11

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define mp make_pair#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;//const double pi = acos(-1);const double eps = 1e-10;//const int dir[[4][2] = { 1,0, -1,0, 0,1, 0,-1 };const int maxn = 100010;const double inf = 1e+20;struct point{    double x, y;};point p[maxn], tmpt[maxn];inline double dist ( point a, point b ){    return sqrt ( ( a.x - b.x )  * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );}bool cmpxy ( point a, point b ){    if ( a.x != b.x )        return a.x < b.x;    else        return a.y < b.y;}bool cmpy ( point a, point b ){    return a.y < b.y;}double closest_pair ( int l, int r ){    double d = inf;    if ( l == r )return d;    if ( l + 1 == r )return dist( p[l], p[r] );    int mid = ( l + r ) >> 1;    double d1 = closest_pair( l, mid );    double d2 = closest_pair( mid + 1, r );    d = min ( d1, d2 );    int k = 0;    for ( int i = l; i <= r; i ++ ){        if ( fabs( p[mid].x - p[i].x ) <= d ){            tmpt[k++] = p[i];        }    }    sort ( tmpt, tmpt + k, cmpy );    for ( int i = 0; i < k; i ++ ){        for ( int j = i +1; j < k && tmpt[j].y - tmpt[i].y < d; j ++ ){            d = min ( d, dist ( tmpt[i], tmpt[j] ) );        }    }    return d;}int main(){    ios::sync_with_stdio( false );    int n;    while ( scanf ( "%d", &n ) && n ){    //while ( cin >> n && n ){        for ( int i = 0; i < n; i ++ ){            scanf ( "%lf%lf", &p[i].x, &p[i].y );            //cin >> p[i].x >> p[i].y;        }        sort ( p, p + n, cmpxy );        printf ( "%0.2f\n", closest_pair( 0, n - 1 ) / 2 );        //cout << fixed << setprecision(2) << closest_pair( 0, n - 1 ) / 2 << endl;    }    return 0;}

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