ACM-计算几何之Raid——poj3714
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 8248 | Accepted: 2460 |
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
Source
/**************************************
***************************************
* Author:Tree *
*From :http://blog.csdn.net/lttree *
* Title : Raid *
*Source: poj 3714 *
* Hint : 计算几何——最近点对 *
***************************************
**************************************/
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
// 不光有点，还要有集合类别标志
struct Point
{
double x,y;
int flag;
}p[1000001];
int arr[1000001];
double Min(double a,double b)
{
return a<b?a:b;
}
// 求两点之间的距离
double dis(Point a,Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
// 根据点横坐标or纵坐标排序
bool cmp_y( int a,int b)
{
return p[a].y<p[b].y;
}
bool cmp_x( Point a,Point b)
{
return a.x<b.x;
}
// 求最近点对
double close_pair( int l,int r )
{
// 如果只剩下两个点，判断是否属于同一集合
if( r==l+1 )
{
if( p[l].flag!=p[r].flag )
return dis( p[l],p[r] );
else return 999999;
}
// 剩下三个点，判断集合类别
else if( r==l+2 )
{
if( p[l].flag==p[l+1].flag )
{
if( p[l].flag==p[l+2].flag ) return 999999;
else return Min( dis(p[l],p[l+2]),dis(p[l+1],p[l+2]) );
}
else
{
if( p[l].flag==p[l+2].flag ) return Min( dis(p[l],p[l+1]),dis(p[l+2],p[l+1]) );
else return Min( dis(p[l],p[l+1]),dis(p[l],p[l+2]) );
}
return Min( dis(p[l],p[r]),Min( dis(p[l],p[l+1]),dis(p[l+1],p[r]) ) );
}
int mid=(l+r)>>1;
double ans=Min(close_pair(l,mid),close_pair(mid+1,r));
int i,j,cnt=0;
for(i=l; i<=r; ++i)
if( p[i].x>=p[mid].x-ans && p[i].x<=p[mid].x+ans )
arr[cnt++]=i;
sort(arr,arr+cnt,cmp_y);
for( i=0; i<cnt ; i++ )
for(j=i+1; j<cnt; j++)
{
if( p[arr[j]].flag != p[arr[i]].flag )
{
if(p[arr[j]].y-p[arr[i]].y>=ans) break;
ans=Min(ans,dis(p[arr[i]],p[arr[j]]));
}
}
return ans;
}
int main()
{
int i,n,t;
scanf("%d",&t);
while( t-- )
{
scanf("%d",&n);
for(i=0;i<n;++i)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
p[i].flag=1;
}
for(;i<n+n;++i)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
p[i].flag=2;
}
sort(p,p+n+n,cmp_x);
printf("%.3lf\n",close_pair(0,n+n-1));
}
return 0;
}