POJ3714 Raid（最近点对、分治）

题目链接：

　　http://poj.org/problem?id=3714

题目描述：

Raid

 

Description

After successive failures in the battles against the Union, the Empire retreated to its last stronghold. Depending on its powerful defense system, the Empire repelled the six waves of Union's attack. After several sleepless nights of thinking, Arthur, General of the Union, noticed that the only weakness of the defense system was its energy supply. The system was charged by N nuclear power stations and breaking down any of them would disable the system.

The general soon started a raid to the stations by N special agents who were paradroped into the stronghold. Unfortunately they failed to land at the expected positions due to the attack by the Empire Air Force. As an experienced general, Arthur soon realized that he needed to rearrange the plan. The first thing he wants to know now is that which agent is the nearest to any power station. Could you, the chief officer, help the general to calculate the minimum distance between an agent and a station?

Input

The first line is a integer T representing the number of test cases.
Each test case begins with an integer N (1 ≤ N ≤ 100000).
The next N lines describe the positions of the stations. Each line consists of two integers X (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the station.
The next following N lines describe the positions of the agents. Each line consists of two integers X (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the agent. 　

Output

For each test case output the minimum distance with precision of three decimal placed in a separate line.

Sample Input

2
4
0 0
0 1
1 0
1 1
2 2
2 3
3 2
3 3
4
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0

Sample Output

1.414
0.000

 

题目大意：

　　有两个点集，求集合不同的点的最小距离

思路：

　　分治，和经典最近点对一样，以x坐标排序，划中线分成两堆，递归求每堆的最近距离d，然后对两堆之间横坐标为 x[mid] - d 和 x[mid] + d 之间的点暴力看是否有比 d 小的距离

　　据说数据比较水可以暴力

　　注意会爆int

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

typedef long long LL;

const LL INF = 1000000000000;
const int N = 100010;

struct Node {
    LL x, y;
    int id;
    Node(LL x = 0, LL y = 0, int id = 0) :x(x), y(y), id(id) {}
    const bool operator < (const Node A) const {
        return x == A.x ? y < A.y : x < A.x;
    }
}no[2 * N];

int n;

double dis(int a, int b) {
    return sqrt((double)((no[a].x - no[b].x)*(no[a].x - no[b].x) + (no[a].y - no[b].y)*(no[a].y - no[b].y)));
}

double solve(int l, int r) {
    if (l == r)return INF;
    int mid = (l + r) >> 1;
    double a = solve(l, mid);
    double b = solve(mid + 1, r);
    double d = min(a, b);
    for (int i = mid; i >= l; --i) {
        if (no[mid].x - no[i].x > d)break;
        for (int j = mid + 1; j <= r; ++j) {
            if (no[j].x - no[i].x > d)break;
            double tmp = dis(i, j);
            if (no[i].id != no[j].id&&tmp < d)d = tmp;
        }
    }
    return d;
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 0; i < n; ++i) {
            scanf("%lld%lld", &no[i].x, &no[i].y);
            no[i].id = 1;
        }
        for (int i = 0; i < n; ++i) {
            scanf("%lld%lld", &no[i + n].x, &no[i + n].y);
            no[i + n].id = 2;
        }
        sort(no, no + 2 * n);
        double ans = solve(0, 2 * n - 1);
        printf("%.3lf\n", ans);
    }
}