POJ 3714 Raid（平面近期点对）

解题思路：

分治法求平面近期点对。点分成两部分，加个标记就好了。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
#include <string.h>
#define LL long long
using namespace std;
const int MAXN = 200000 + 10;
const double INF = 1e100;
struct Point
{
    double x, y;
    int flag;
}P[MAXN];
int N;
Point vec[MAXN];
bool cmp_x(Point a, Point b)
{
    return a.x < b.x;
}
bool cmp_y(Point a, Point b)
{
    return a.y < b.y;
}
double dis(Point a, Point b)
{
    double dx = a.x - b.x;
    double dy = a.y - b.y;
    return sqrt(dx * dx + dy * dy);
}
double solve(Point *a, int l, int r)
{
    if(l == r) return INF;
    if(l + 1 == r)
    {
        if(P[l].flag == P[r].flag)
            return INF;
        return dis(P[l], P[r]);
    }
    int m = (l + r) >> 1;
    double d = solve(a, l, m);
    d = min(d, solve(a, m + 1, r));
    int sz = 0;
    for(int i=l;i<=r;i++)
    {
        if(fabs(P[i].x - P[m].x) <= d)
            vec[sz++] = P[i];
    }
    sort(vec, vec + sz, cmp_y);
    for(int i=0;i<sz;i++)
    {
        for(int j=i+1;j<sz;j++)
        {
            if(fabs(vec[i].y - vec[j].y) >= d)
                break;
            if(vec[i].flag != vec[j].flag)
            {
                double rs = dis(vec[i], vec[j]);
                if(rs < d) d = rs;
            }
        }
    }
    return d;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &N);
        for(int i=0;i<N;i++)
        {
            scanf("%lf%lf", &P[i].x, &P[i].y);
            P[i].flag = 0;
        }
        for(int i=0;i<N;i++)
        {
            scanf("%lf%lf", &P[i + N].x, &P[i + N].y);
            P[i + N]. flag = 1;
        }
        N <<= 1;
        sort(P, P + N, cmp_x);
        double ans = solve(P, 0, N - 1);
        printf("%.3f\n", ans);
    }
    return 0;
}