题目链接：
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.
First line of the input contains six integers ax, ay, bx, by, tx and ty (0 ≤ ax, ay, bx, by, tx, ty ≤ 10^9) — 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 ≤ 10^9) — position of the i-th bottle.
It's guaranteed that positions of Adil, Bera, recycling bin and all bottles are distinct.
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 
.
3 1 1 2 0 0
3
1 1
2 1
2 3
11.084259940083
5 0 4 2 2 0
5
5 2
3 0
5 5
3 5
3 3
33.121375178000
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.
题意：
给出两个人的初始位置和垃圾桶的位置以及那些瓶子的位置,每个人可以按上面给的步骤进行,问最后两个人走的最少的距离是多少;
思路：
把每个瓶子到垃圾桶和两个人的距离都求出来,再贪心找到最小距离;
最小距离的基础是每个瓶子到垃圾桶距离和的2倍即sum=Σdis[i],然后再选择一个人或者两个人去捡一个瓶子使sum-dis[i]+disa[i]-dis[j]+disb[j]最小,当然这是两个人都去捡的情况,还有就是一个人去捡的情况sum-dis[i]+(disa[i]ordisb[i]);
反正就是找最小的值啦;
本来思路是对的，比赛的时候代码有个地方是b的我直接把a的复制过去忘记改成b了,最后挂在了st上了;
AC代码：
#include <bits/stdc++.h> /*#include <iostream> #include <queue> #include <cmath> #include <cstring> #include <algorithm> #include <cstdio> */ using namespace std; #define Riep(n) for(int i=1;i<=n;i++) #define Riop(n) for(int i=0;i<n;i++) #define Rjep(n) for(int j=1;j<=n;j++) #define Rjop(n) for(int j=0;j<n;j++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=0x3f3f3f3f; const int N=1e5+25; double fun(double xx,double yy,double fx,double fy) { return sqrt((xx-fx)*(xx-fx)+(yy-fy)*(yy-fy)); } double ax,ay,bx,by,tx,ty,x[N],y[N],dis[N],disa[N],disb[N]; int n; int main() { scanf("%lf%lf%lf%lf%lf%lf",&ax,&ay,&bx,&by,&tx,&ty); scanf("%d",&n); double ans=0; Riep(n) { scanf("%lf%lf",&x[i],&y[i]); dis[i]=fun(x[i],y[i],tx,ty); disa[i]=fun(x[i],y[i],ax,ay); disb[i]=fun(x[i],y[i],bx,by); ans+=2*dis[i]; } disa[0]=0; disb[0]=0; dis[0]=0; double mmina=999999999999999,mminb=99999999999999,s=ans; int posa=0,posb=0; for(int i=1;i<=n;i++) { if(disa[i]-dis[i]<mmina) { mmina=disa[i]-dis[i]; posa=i; } if(disb[i]-dis[i]<mminb) { mminb=disb[i]-dis[i]; posb=i; } } if(mmina<0&&mminb<0) { if(posa==posb) { for(int i=0;i<=n;i++) { if(i!=posa) s=min(s,ans-dis[posa]+disa[posa]-dis[i]+disb[i]); } for(int i=0;i<=n;i++) { if(i!=posb) s=min(s,ans-dis[posb]+disb[posb]-dis[i]+disa[i]); } } else { s=ans+mmina+mminb; } } else { if(mmina<mminb) { s=ans-dis[posa]+disa[posa]; } else { s=ans-dis[posb]+disb[posb]; } } printf("%.10lf ",s); return 0; }