Codeforces 671A Recycling Bottles 【dp】

题目链接：Codeforces 671A Recycling Bottles

A. Recycling Bottles

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

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.

题意：给定n个垃圾堆的坐标，已知两个人A、B以及垃圾回收站的坐标。要求每次只能让一个人工作，且每处理一个垃圾堆必须把垃圾放进回收站。现在要把所有垃圾扔进回收站，问你A、B两人所走的最小距离。

思路：dp[i][j][k][l]表示处理前i个点且用第j个人(0表示A，1表示B)走第i个点，A的位置状态为k，B的位置状态为l的最优解。位置状态为0时表示在回收站，反之在起点。

因为在处理某个点的过程中，从起点开始走可能不是最优的，这时需要从回收站出发才是最优的。

用dp[i][2][0][0]表示处理前i个点两人均在回收站的状态，因为最优解可能产生——先到第i+1个点，再由回收站到达前面的点。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
const int MAXN = 1e5 + 10;
double dp[MAXN][3][2][2];
double d1[MAXN], d2[MAXN], d3[MAXN];
double dis(int x1, int y1, int x2, int y2) {
return sqrt(1LL * (x1 - x2) * (x1 - x2) * 1.0 + 1LL * (y1 - y2) * (y1 - y2));
}
void Solve(double &x, double y) {
if(y == -1) return ;
if(x == -1) x = y;
else x = min(x, y);
}
int main()
{
int ax, ay, bx, by, tx, ty;
while(scanf("%d%d%d%d%d%d", &ax, &ay, &bx, &by, &tx, &ty) != EOF) {
int n; scanf("%d", &n);
for(int i = 1; i <= n; i++) {
int x, y; scanf("%d%d", &x, &y);
d1[i] = dis(ax, ay, x, y);
d2[i] = dis(bx, by, x, y);
d3[i] = dis(tx, ty, x, y);
dp[i][0][1][0] = dp[i][0][1][1] = dp[i][1][0][1] = dp[i][1][1][1] = -1;
dp[i][2][0][0] = -1;
}
dp[1][2][0][0] = 2 * d3[1];
dp[1][0][1][0] = d1[1] + d3[1];
dp[1][1][0][1] = d2[1] + d3[1];
for(int i = 1; i < n; i++) {
Solve(dp[i+1][2][0][0], dp[i][2][0][0] + 2 * d3[i+1]);
Solve(dp[i+1][0][1][0], dp[i][2][0][0] + d1[i+1] + d3[i+1]);
Solve(dp[i+1][1][0][1], dp[i][2][0][0] + d2[i+1] + d3[i+1]);

if(dp[i][0][1][0] != -1) {
Solve(dp[i+1][0][1][0], dp[i][0][1][0] + 2 * d3[i+1]);
Solve(dp[i+1][1][1][1], dp[i][0][1][0] + d2[i+1] + d3[i+1]);
}

if(dp[i][0][1][1] != -1) {
Solve(dp[i+1][0][1][1], dp[i][0][1][1] + 2 * d3[i+1]);
Solve(dp[i+1][1][1][1], dp[i][0][1][1] + 2 * d3[i+1]);
}

if(dp[i][1][0][1] != -1) {
Solve(dp[i+1][0][1][1], dp[i][1][0][1] + d1[i+1] + d3[i+1]);
Solve(dp[i+1][1][0][1], dp[i][1][0][1] + 2 * d3[i+1]);
}

if(dp[i][1][1][1] != -1) {
Solve(dp[i+1][0][1][1], dp[i][1][1][1] + 2 * d3[i+1]);
Solve(dp[i+1][1][1][1], dp[i][1][1][1] + 2 * d3[i+1]);
}
}
double ans = -1;
//printf("%.8lf %.8lf %.8lf %.8lf\n", dp
[0][1][0], dp
[0][1][1], dp
[1][0][1], dp
[1][1][1]);
Solve(ans, dp
[0][1][0]); Solve(ans, dp
[0][1][1]);
Solve(ans, dp
[1][0][1]); Solve(ans, dp
[1][1][1]);
printf("%.8lf\n", ans);
}
return 0;
}