POJ 3714 Raid（变种最近点对问题：分治+剪枝）

Raid

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 11728 Accepted: 3494

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

240 00 11 01 12 22 33 23 340 00 00 00 00 00 00 00 0

Sample Output

1.4140.000
/*分治剪枝，直接把代理和车站放在同一数组中，只是在初始化时给予不同的flag直接套用HDU1007的模板即可，在判断最小距离时先判断flag是否不等即可*/#include<iostream>#include<cstring>#include<vector>#include<algorithm>#include<cmath>#include<cstdio>using namespace std;const double eps = 1e-8;const double PI = acos(-1.0);#define INF 0x3f3f3f3f#define maxn 1000100int sgn(double x){if (fabs(x) < eps)return 0;if (x < 0)return -1;else return 1;}struct Point{double x, y;int flag;Point() {}Point(double _x, double _y){x = _x; y = _y;}Point operator -(const Point &b)const{return Point(x - b.x, y - b.y);}//叉积double operator ^(const Point &b)const{return x*b.y - y*b.x;}//点积double operator *(const Point &b)const{return x*b.x + y*b.y;}//绕原点旋转角度B（弧度值），后x,y的变化void transXY(double B){double tx = x, ty = y;x = tx*cos(B) - ty*sin(B);y = tx*sin(B) + ty*cos(B);}}p[maxn], t[maxn];double dist(Point a, Point b){if (a.flag != b.flag)return sqrt((a - b)*(a - b));elsereturn INF;}bool cmpx(const Point& a, const Point& b){return a.x<b.x || (a.x == b.x && a.y<b.y);}bool cmpy(const Point& a, const Point& b){return a.y<b.y || (a.y == b.y && a.x<b.x);}double closest(int left, int right){double d = INF;if (left == right)return d;if (left + 1 == right)return dist(p[left], p[right]);int mid = (left + right) / 2;double d1 = closest(left, mid);double d2 = closest(mid + 1, right);d = min(d1, d2);int k = 0;for (int i = left; i <= right; i++)if (fabs(p[i].x - p[mid].x) <= d)t[k++] = p[i];sort(t, t + k, cmpy);for (int i = 0; i < k; i++)for (int j = i + 1; j < k&&t[j].y - t[i].y < d; j++)d = min(d, dist(t[i], t[j]));return d;}int main(){int n,T;scanf("%d", &T);while ( T--){scanf("%d", &n);double x, y;for (int i = 0; i<n; i++){scanf("%lf%lf", &x, &y);p[i] = Point(x, y);p[i].flag = 0;}for (int i = n; i<2*n; i++){scanf("%lf%lf", &x, &y);p[i] = Point(x, y);p[i].flag = 1;}sort(p, p + 2*n, cmpx);printf("%.3f\n", closest(0, 2*n - 1));}return 0;}