Recycling Bottles

最新推荐文章于 2025-11-04 09:10:26 发布

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本文探讨了在一个二维坐标平面上，两名收集者如何高效地从不同起点出发，收集多个瓶子并投放到指定回收箱中，以使得行走总距离最短的问题。

It was recycling day in Kekoland. To celebrate it Adil and Bera went to Central Perk where they can take bottles from the ground and put them into a recycling bin.

We can think Central Perk as coordinate plane. There are n bottles on the ground, the i-th bottle is located at position (xi, yi). Both Adil and Bera can carry only one bottle at once each.

For both Adil and Bera the process looks as follows:

Choose to stop or to continue to collect bottles.
If the choice was to continue then choose some bottle and walk towards it.
Pick this bottle and walk to the recycling bin.
Go to step 1.

Adil and Bera may move independently. They are allowed to pick bottles simultaneously, all bottles may be picked by any of the two, it's allowed that one of them stays still while the other one continues to pick bottles.

They want to organize the process such that the total distance they walk (the sum of distance walked by Adil and distance walked by Bera) is minimum possible. Of course, at the end all bottles should lie in the recycling bin.

Input

First line of the input contains six integers ax, ay, bx, by, tx and ty (0 ≤ ax, ay, bx, by, tx, ty ≤ 109) — initial positions of Adil, Bera and recycling bin respectively.

The second line contains a single integer n (1 ≤ n ≤ 100 000) — the number of bottles on the ground.

Then follow n lines, each of them contains two integers xi and yi (0 ≤ xi, yi ≤ 109) — position of the i-th bottle.

It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.

Output

Print one real number — the minimum possible total distance Adil and Bera need to walk in order to put all bottles into recycling bin. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if $\text{[math]}$ .

Examples

input
3 1 1 2 0 0
3
1 1
2 1
2 3

output
11.084259940083

input
5 0 4 2 2 0
5
5 2
3 0
5 5
3 5
3 3

output
33.121375178000

Note

Consider the first sample.

Adil will use the following path: $\text{[math]}$ .

Bera will use the following path: $\text{[math]}$ .

Adil's path will be $\text{[math]}$ units long, while Bera's path will be $\text{[math]}$ units long.

题意：给定2个人的坐标和垃圾桶的坐标，以及N的垃圾的坐标，问将所以垃圾放到垃圾桶的距离和最小是多少。可以一个人不捡。

题解：当A和B都到垃圾桶时，谁去捡都是一样的，来回都是垃圾到垃圾桶的距离，我们要考虑的是A去捡的第一个垃圾以及B去捡的第一个垃圾。

我们用dis[0][i]表示A的初始位置到第i个垃圾的距离，dis[1][i]表示B的初始位置到第i个垃圾的距离，dis[2][i]表示垃圾桶到第i个垃圾的位置。

我们可以先把从垃圾桶出发的距离和tot = sum(dis[2][1]....dis[2][n])

这样，如果A去捡第i个垃圾，那么就可以在tot的基础上减少一次dis[2][i]但是要增加一次dis[0][i],然后B去捡第j(j!=i)个垃圾，可以再减少一次dis[2][j]然后增加一次dis[1][j].当然此时B可以不去捡垃圾。

然后我们再反的来一次，让B先去捡，考虑A捡哪个，或者不捡，比较两次花费时间即可。

#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define fur(i,a,b) for(int i=(a);i<=(b);i++)
#define furr(i,a,b) for(int i=(a);i>=(b);i--)
#define cl(a) memset((a),0,sizeof(a))
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define sc(x) scanf("%d",&x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair <int, int> pii;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const int mod=1000000007;
const double pi=acos(-1);
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
inline void gn(double&x){double t;scanf("%lf",&t);x=t;}
inline void gn(long double&x){double t;scanf("%lf",&t);x=t;}
int gcd(int a,int b){return a? gcd(b%a,a):b;}
ll powmod(ll a,ll x,ll mod){ll t=1ll;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
struct Node{
	int x,y;
}pa,pb,p,tmp;
double	getdis(Node a,Node b){
	return sqrt((a.x-b.x)*1.0*(a.x-b.x)+(a.y-b.y)*1.0*(a.y-b.y));
}
double dis[3][1000005];
int main(){
	//freopen("E:\\data.in","r",stdin);
	gn(pa.x);gn(pa.y);gn(pb.x);gn(pb.y);gn(p.x);gn(p.y);
	int n;
	dis[0][0] = dis [1][0] = dis[2][0] = 0;
	double tot = 0;
	gn(n);
	fur(i,1,n){
		gn(tmp.x);gn(tmp.y);
		dis[0][i] = getdis(tmp,pa);
		dis[1][i] = getdis(tmp,pb);
		dis[2][i] = getdis(tmp,p);
		tot+=(dis[2][i]*2);
	}
	
	
	double add1 = 1e18;
	int id = 0;
	fur(i,1,n)
		if(dis[0][i]-dis[2][i]<add1){
			add1 = dis[0][i]-dis[2][i];
			id = i;
		}
	double add2 = dis[1][0]-dis[2][0];
	fur(i,1,n)
		if(i == id)continue;	//两人第一个捡的不能相同 
		else if(dis[1][i]-dis[2][i]<add2)
			add2 = dis[1][i]-dis[2][i];
	
	double plus = add1+add2;

	add1 = 1e18;
	id = 0;
	fur(i,1,n)
		if(dis[1][i]-dis[2][i]<add1){
			add1 = dis[1][i]-dis[2][i];
			id = i;
		}
	add2 = dis[0][0]-dis[2][0];
	fur(i,0,n)
		if(i == id)continue;
		else if(dis[0][i]-dis[2][i]<add2)
			add2 = dis[0][i]-dis[2][i];


	if(add1+add2<plus)
		plus = add1+add2;
		
	printf("%lf\n",tot+plus);
	return 0;
}