[POJ] 3714 -> Raid

Raid

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 8937		Accepted: 2694

DescriptionAfter 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, Generalof the Union, noticed that the only weakness of the defense system was its energy supply. The system was charged byN 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, Arthursoon 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?InputThe 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 integersX (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 integersX (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the agent. 　OutputFor 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

解题思路：

最近点对问题，可以用《算法导论》上的分治法求解

这里给一种简单的暴力+剪枝算法

Code:

#include <iostream>#include <cmath>#include <algorithm>using namespace std;struct Point{double x, y;bool isStation;};inline double dist(Point& pt1, Point& pt2){double dx = pt1.x - pt2.x;double dy = pt1.y - pt2.y;return sqrt(dx*dx + dy*dy);}bool cmp(Point& pt1, Point& pt2){if (pt1.x == pt2.x) return pt1.y < pt2.y;return pt1.x < pt2.x;}Point points[200004];int main(){int t, n;double x, y;cin >> t;while (t--){cin >> n;for (int i = 0; i < 2 * n; i++){cin >> points[i].x >> points[i].y;if (i < n) points[i].isStation = true;else points[i].isStation = false;}sort(points, points+2*n, cmp);double min = dist(points[0], points);for (int i = 0; i < 2 * n; ++i){for (int j = i + 1; j < 2 * n; ++j){if (points[i].isStation == points[j].isStation) continue;double distance = dist(points[i], points[j]);if (distance < min) min = distance;if ((points[j].x - points[i].x) >= min) break;}}printf("%.3f\n", min);}return 0;}