codeforces 671A Recycling Bottles

A. 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.

引用自：点击打开链接

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;

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <algorithm>

using namespace std;

const int maxn = 100000;
int n;
double ax, ay, bx, by, tx, ty; //刚开始就用double，用int会越界
double dis[maxn + 10], disa[maxn + 10], disb[maxn + 10];

double getdis(double x1, double y1, double x2, double y2) {
	return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

int main()
{
	scanf("%lf%lf%lf%lf%lf%lf", &ax, &ay, &bx, &by, &tx, &ty);
	scanf("%d", &n);
	double ans = 0, s;
	for (int i = 1; i <= n; i++) {
		double x, y;
		scanf("%lf%lf", &x, &y);
		dis[i] = getdis(x, y, tx, ty);
		disa[i] = getdis(x, y, ax, ay);
		disb[i] = getdis(x, y, bx, by);
		ans += dis[i] * 2;
	}
	s = ans;
	double minna = 999999999999, minnb = 999999999999;
	int posa = 1, posb = 1;
	for (int i = 1; i <= n; i++) {
		if (disa[i] - dis[i] < minna) {
			minna = disa[i] - dis[i];
			posa = i;
		}
		if (disb[i] - dis[i] < minnb) {
			minnb = disb[i] - dis[i];
			posb = i;
		}
	}
	dis[0] = disa[0] = disb[0] = 0;
	//当minna < 0时disa[i] < dis[i]，也就是说第一次去捡的时候
	//disa[i] + dis[i] < dis[i] * 2，即从A的位置过去捡起来再丢进垃圾桶
	//比以后从垃圾桶出发折返一次路径要短，B同理
	if (minna < 0 && minnb < 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 = s + minna + minnb;
		}
	}
	else {
		//这时候说明有，假如是minna >= 0
		//disa[i] >= dis[i]，也就是第一次A的位置没有对最后结果做出积极的作用
		//但是第一步再怎么样都不能不捡东西直接去垃圾桶，不然最后总路径肯定更长
		//无奈之下只好只让A或者B中一个人一直捡完，另一个人打酱油不动
		//选捡第一个的时候走的路径尽量短的那个人去捡
		if (minna < minnb) {
			s = s + minna;
		}
		else {
			s = s + minnb;
		}
	}
	printf("%.12f\n", s);
	return 0;
}