time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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:
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 .
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
【题解】
这道题求的是两个人总的最短的路程。
而且每次只能拿一个瓶子；
那么。如果两个人都到了垃圾桶的位置；
接下来谁去拿都一样了->完全可以一个人就站在垃圾桶的位置不动。->因为并不是求最短时间！
这样就转换成一个人的问题了；
同时每个垃圾到垃圾桶的距离也是固定的。
预处理一下即可；
设
dis[0][i]表示垃圾桶到第i个垃圾的距离；
dis[1][i]表示第一个人到第i个垃圾的距离;
dis[2][i]表示第二个人到第i个垃圾的距离;
则一开始预处理出sum=∑2*dis[0][i]
然后分类讨论;
1.第一个人捡起某个垃圾扔到垃圾桶，此后都由他捡。另外一个人至始至终都不动；
2.第二个人捡起某个垃圾扔到垃圾桶，此后都由他捡。另外一个人至始至终都不动；
3.两个人人都捡起一个垃圾(都选离自己最近的)，扔到垃圾桶的位置。然后都由一个人捡。另外一个人站在垃圾桶的位置不动；
这里一个人的情况不会出现什么特殊的。
两个人的话，可能选择的是同一个垃圾X；
则处理出离两个人第二远的垃圾；
分别试试让第一个人捡这个X，第二个人捡离自己第二远的垃圾
或者是第一个人捡离自己第二远的垃圾，第二个人捡这个X；
捡垃圾Y的过程
可以表述为
sum-=dis[0][Y];
sum+=dis[t][Y];
t表示的是第几个人;
第一个人捡则加上dis[1][Y];
第二个人捡则加上dis[2][Y]
#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#define LL long long
using namespace std;
const int MAXN = 1e5 + 100;
const double INF = 1e18;
int ax, ay, bx, by, tx, ty, n;
int x[MAXN], y[MAXN];
double dis[3][MAXN];
void input(int &r)
{
r = 0;
char t = getchar();
while (!isdigit(t)) t = getchar();
int sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
double sqr(double x)
{
return x*x;
}
double get_dis(double a0, double b0, double a1, double b1)
{
return sqrt(sqr(a0 - a1) + sqr(b0 - b1));
}
int main()
{
//freopen("F:\\rush.txt", "r", stdin);
input(ax); input(ay); input(bx); input(by); input(tx); input(ty);
input(n);
for (int i = 1; i <= n; i++)
input(x[i]), input(y[i]);
for (int i = 1; i <= n; i++)
{
dis[0][i] = get_dis(tx, ty, x[i], y[i]);
dis[1][i] = get_dis(ax, ay, x[i], y[i]);
dis[2][i] = get_dis(bx, by, x[i], y[i]);
}
double sum = 0, fans;
for (int i = 1; i <= n; i++)
sum += 2 * dis[0][i];
//第一个人捡(一定选) 第二个人不捡
double misum1 = INF;
for (int i = 1; i <= n; i++)
{
double tempsum = sum;
tempsum -= dis[0][i];
tempsum += dis[1][i];
misum1 = min(misum1, tempsum);
}
fans = misum1;
//第二个人捡(一定选) 第一个人不捡
misum1 = INF;
for (int i = 1; i <= n; i++)
{
double tempsum = sum;
tempsum -= dis[0][i];
tempsum += dis[2][i];
misum1 = min(misum1, tempsum);
}
fans = min(fans, misum1);
//第一个个人捡，第二个人也捡
if (n > 1)
{
int num1;
misum1 = INF;
for (int i = 1; i <= n; i++)
{
double tempsum = sum;
tempsum -= dis[0][i];
tempsum += dis[1][i];
if (tempsum < misum1)
{
num1 = i;
misum1 = tempsum;
}
}
int num2;
double misum2 = INF;
for (int i = 1; i <= n; i++)
{
double tempsum = sum;
tempsum -= dis[0][i];
tempsum += dis[2][i];
if (tempsum < misum2)
{
num2 = i;
misum2 = tempsum;
}
}
if (num1 != num2)//捡的是不同的垃圾则可直接求得答案
{
double tempsum = sum;
tempsum -= dis[0][num1];
tempsum -= dis[0][num2];
tempsum += dis[1][num1];
tempsum += dis[2][num2];
fans = min(fans, tempsum);
}
else
{
int tnum1;
misum1 = INF;
for (int i = 1; i <= n; i++)
if (i != num1)
{
double tempsum = sum;
tempsum -= dis[0][i];
tempsum += dis[1][i];
if (tempsum < misum1)
{
tnum1 = i;
misum1 = tempsum;
}
}
int tnum2;
double misum2 = INF;
for (int i = 1; i <= n; i++)
if (i != num1)
{
double tempsum = sum;
tempsum -= dis[0][i];
tempsum += dis[2][i];
if (tempsum < misum2)
{
tnum2 = i;
misum2 = tempsum;
}
}
double tempsum = sum;
tempsum -= dis[0][num1];
tempsum -= dis[0][tnum2];
tempsum += dis[1][num1];
tempsum += dis[2][tnum2];
fans = min(fans, tempsum);
tempsum = sum;
tempsum -= dis[0][num1];
tempsum -= dis[0][tnum1];
tempsum += dis[2][num1];
tempsum += dis[1][tnum1];
fans = min(fans, tempsum);
}
}
printf("%.12lf\n", fans);
return 0;
}