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POJ 3714 Raid 最近点对

 
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求最近点对,只不过这两个点需要属于不同的集合,那么就给两个集合的点分别标记一个id号,在计算时,两个集合合并起来,并排序,递归求解,只不过,求两点距离时,如果id号是同一集合的,直接返回一个很大的数就行了,这样就跟求一个集合的最近点对没什么区别了。

/*
ID: sdj22251
PROG: calfflac
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAX 2000000000
#define LOCA
using namespace std;
struct node
{
    double x;
    double y;
    int id;
}p[200005],p1[100005],p2[100005];
bool cmp(node x, node y)
{
    return x.x < y.x;
}
double dist(node x, node y)
{
    if(x.id != y.id)
    return sqrt((x.x - y.x) * (x.x - y.x) + (x.y - y.y) * (x.y - y.y));
    else return 2100000000.0; // id号相同肯定不行,直接返回一个很大的数
}
double Divide_conquer(int low, int high)
{
    if(high == low) return 2100000000.0; 
    if(high - low == 1) return dist(p[low], p[high]);
    int mid = (low + high) / 2;
    double d1 = Divide_conquer(low, mid);
    double d2 = Divide_conquer(mid + 1, high);
    double d = min(d1, d2);
    int i, j, cnt1 = 0, cnt2 = 0;
    for(i = mid; i >= low; i--)
    {
        if(p[mid].x - p[i].x < d)
        {
            p1[cnt1++] = p[i];
        }
    }
    for(i = mid + 1; i <= high; i++)
    {
        if(p[i].x - p[mid].x < d)
        {
            p2[cnt2++] = p[i];
        }
    }
    for(i = 0; i < cnt1; i++)
    {
        for(j = 0; j < cnt2; j++)
        {
            d = min(d, dist(p1[i], p2[j]));
        }
    }
    return d;
}
int main()
{
#ifdef LOCAL
    freopen("calfflac.in","r",stdin);
    freopen("calfflac.out","w",stdout);
#endif
    int t, n, i, j;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        for(i = 0; i < n; i++)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
            p[i].id = 1;
        }
        for(i = n; i < 2 * n; i++)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
            p[i].id = 2;
        }
        n = n * 2;
        sort(p, p + n, cmp);
        printf("%.3f\n", Divide_conquer(0, n - 1));
    }
	return 0;
}


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