Codeforces Recycling Bottles 模拟

C. Recycling Bottles

time limit per test：

2 seconds

memory limit per test：

256 megabytes

input：

standard input

output：

standard output

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:

1. Choose to stop or to continue to collect bottles.
2. If the choice was to continue then choose some bottle and walk towards it.
3. Pick this bottle and walk to the recycling bin.
4. 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 .

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: .

Bera will use the following path: .

Adil's path will be  units long, while Bera's path will be  units long.

题目链接：http://codeforces.com/contest/672/problem/C

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题意：一个广场有n个瓶子，有a，b两个人和一个垃圾桶。捡了瓶子后马上丢到垃圾桶，求两个人的最短路程。

思路：第一次捡瓶子要走的距离就是人到瓶子的距离+瓶子到垃圾桶的距离，第二次捡瓶子要走的距离就是另外一个人到瓶子的距离+瓶子到垃圾桶的距离（或者另外一个人不捡瓶子）。捡其他瓶子要走的距离是垃圾桶到瓶子的距离*2。做差值进行排序。sign1，sign2表示一个人差值最大的两个值。sign3，sign4表示另外一个人差值最大的两个值。总共有5种情况表示人到瓶子的距离+瓶子到垃圾桶的距离（1）sign1；（2）sign3;（3）sign1，sign3；（4）sign1，sign4；（5）sign2，sign3；其中第3种情况要两个瓶子不是同一个。

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代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int MAXN = 1e5+, mod = 1e9 + , inf = 0x3f3f3f3f;
typedef long long ll;
const ll INF = (1ll<<);
struct gg
{
    double d;
    int i;
} sub1[],sub2[];
int cmp(gg a,gg b)
{
    return a.d>b.d;
}
int main()
{
    int i;
    double ax,ay,bx,by,tx,ty;
    int n;
    double x,y;
    double ans=;
    scanf("%lf%lf%lf%lf%lf%lf",&ax,&ay,&bx,&by,&tx,&ty);
    scanf("%d",&n);
    for(i=; i<n; i++)
    {
        scanf("%lf%lf",&x,&y);
        double sign1=sqrt((x-ax)*(x-ax)+(y-ay)*(y-ay))+sqrt((x-tx)*(x-tx)+(y-ty)*(y-ty));
        double sign2=sqrt((x-bx)*(x-bx)+(y-by)*(y-by))+sqrt((x-tx)*(x-tx)+(y-ty)*(y-ty));
        double d=2.0*sqrt((x-tx)*(x-tx)+(y-ty)*(y-ty));
        sub1[i].d=d-sign1;
        sub2[i].d=d-sign2;
        sub1[i].i=sub2[i].i=i;
        ans+=d;
    }
    sort(sub1,sub1+n,cmp);
    sort(sub2,sub2+n,cmp);
    double sign1=sub1[].d,sign2=sub1[].d,sign3=sub2[].d,sign4=sub2[].d;
    int sign1_i=sub1[].i,sign2_i=sub1[].i,sign3_i=sub2[].i,sign4_i=sub2[].i;
    double sign,Min=1e15;
    sign=ans-sign1;
    if(sign<Min) Min=sign;
    sign=ans-sign3;
    if(sign<Min) Min=sign;
    if(sign1_i!=sign3_i)
    {
        sign=ans-sign1-sign3;
        if(sign<Min) Min=sign;
    }
    sign=ans-sign1-sign4;
    if(sign<Min) Min=sign;
    sign=ans-sign2-sign3;
    if(sign<Min) Min=sign;
    printf("%.12f\n",Min);
    return ;
}