poj 3714 Raid (类型不同点间的最近点对)

 http://poj.org/problem?id=3714

 

 

#include "stdio.h"
#include "cmath"
#include "iostream"
#include "algorithm"
using namespace std;

struct point
{
    long long x, y;
    bool flag;
};
point p[200003];
point tp[200003];

bool cmp_x(const point & a, const point & b) { return a.x < b.x; }

bool cmp_y(const point & a, const point & b) { return a.y < b.y; }

double min(double a, double b) { return a < b ? a : b; }

double dis(const point & a, const point & b)
{
    if(a.flag == b.flag) return 1e12;  //将类型相同的点之间的距离设为无穷大

    return sqrt((double)((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)));
}

double solve(int lf, int rt, int n)  //left, right
{
    if(lf == rt) return 1e12;  //只有一个点时，距离设为无穷大
    if(rt - lf == 1)
        return dis(p[lf], p[rt]);

    int mid = (lf + rt) >> 1;
    double ans = min(solve(lf, mid, n), solve(mid + 1, rt, n));

    int lp = 0, i, j;
    for(i = mid; i >= lf && p[i].x - p[mid].x < ans; i--)
    {
        tp[lp].x = p[i].x;
        tp[lp].y = p[i].y;
        tp[lp++].flag = p[i].flag;
    }

    for(i = mid + 1; i <= rt && p[i].x - p[mid].x < ans; i++)
    {
        tp[lp].x = p[i].x;
        tp[lp].y = p[i].y;
        tp[lp++].flag = p[i].flag;
    }

    sort(tp, tp + lp, cmp_y);

    for(i = 0; i < lp; i++)
        for(j = i + 1; j - i < 7 && j < lp; j++)
            ans = min(ans, dis(tp[i], tp[j]));

    return ans;
}

int main()
{
    int n, i, t;
    cin >> t;
    while(t--)
    {
        scanf("%d", &n);
        for(i = 1; i <= n; i++)
        {
            p[i].flag = true;
            scanf("%I64d%I64d", &p[i].x, &p[i].y);
        }

        for(i = n + 1; i <= 2 * n; i++)
        {
            p[i].flag = false;
            scanf("%I64d%I64d", &p[i].x, &p[i].y);
        }

        n += n;
        sort(p + 1, p + n + 1, cmp_x);
        printf("%.3lf\n", solve(1, n, n));
    }
    return 0;
}