【POJ 3714】 Raid

【题目链接】

http://poj.org/problem?id=3714

【算法】

分治求平面最近点对

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 100010
const double INF = 1e10;

int T,i,n;

struct info
{
        double x,y;
        int opt;
} a[MAXN<<1];

inline bool cmpx(info a,info b)
{
        return a.x != b.x ? a.x < b.x : a.y < b.y;
}
inline bool cmpy(info a,info b)
{
        return a.y < b.y;
}
double dist(info a,info b)
{
        return (a.opt != b.opt) ? sqrt(abs(a.x - b.x) * abs(a.x - b.x) + abs(a.y - b.y) * abs(a.y - b.y)) : INF;
}
inline double Closest_Pair(int l,int r)
{
        int i,j,mid,len = 0;
        static info s[MAXN];
        double d;
        if (l == r) return INF;
        if (l + 1 == r) return dist(a[l],a[r]);
        mid = (l + r) >> 1;
        d = min(Closest_Pair(l,mid),Closest_Pair(mid+1,r));
        for (i = l; i <= r; i++)
        {
                if (abs(a[mid].x - a[i].x) <= d) s[++len] = a[i];
        }
        sort(s+1,s+len+1,cmpy);
        for (i = 1; i <= len; i++)
        {
                for (j = i + 1; j <= len && s[j].y - s[i].y <= d; j++)
                {
                        d = min(d,dist(s[i],s[j]));
                }
        }
        return d;
}

int main()
{

        scanf("%d",&T);
        while (T--)
        {
                scanf("%d",&n);
                for (i = 1; i <= n; i++)
                {
                        scanf("%lf%lf",&a[i].x,&a[i].y);
                        a[i].opt = 1;
                }
                for (i = n + 1; i <= 2 * n; i++)
                {
                        scanf("%lf%lf",&a[i].x,&a[i].y);
                        a[i].opt = 2;
                }
                sort(a+1,a+2*n+1,cmpx);
                printf("%.3lf\n",Closest_Pair(1,n<<1));
        }

        return 0;

}

原文地址：https://www.cnblogs.com/evenbao/p/9240617.html