2019 GDUT Rating Contest II : Problem F. Teleportation

题面：

Problem F. Teleportation

Input file: standard input

Output file: standard output

Time limit: 15 second

Memory limit: 1024 megabytes

 

One of the farming chores Farmer John dislikes the most is hauling around lots of cow manure. In order to streamline this process, he comes up with a brilliant invention: the manure teleporter! Instead of hauling manure between two points in a cart behind his tractor, he can use the manure teleporter to instantly transport manure from one location to another.

Farmer John’s farm is built along a single long straight road, so any location on his farm can be described simply using its position along this road (effectively a point on the number line). A teleporter is described by two numbers x and y, where manure brought to location x can be instantly transported to location y, or vice versa.

Farmer John wants to transport manure from location a to location b, and he has built a teleporter that might be helpful during this process (of course, he doesn’t need to use the teleporter if it doesn’t help). Please help him determine the minimum amount of total distance he needs to haul the manure using his tractor.

 

Input

The first and only line of input contains four space-separated integers: a and b,describing the start and end locations, followed by x and y, describing the teleporter. All positions are integers in the range 0...100, and they are not necessarily distinct from each-other.

 

Output

Print a single integer giving the minimum distance Farmer John needs to haul manure in his tractor.

 

Example

Input

3 10 8 2

Output

3

 

Note

In this example, the best strategy is to haul the manure from position 3 to position 2, teleport it to position 8, then haul it to position 10. The total distance requiring the tractor is therefore 1 + 2 = 3.

 

题目描述：

农夫最不喜欢清理牛粪，于是发明了牛粪传送器：可以使牛粪从一个地方直接传送到另一个地方，而不是从一个地方用推车搬运牛粪到另一个地方。农夫的农场可以想象为一条长直道。现在，给出农夫要把牛粪从a运送到b，还有传送器可以从x传送到y或者y传送到x，问农夫要用推车搬运牛粪的最少距离。

 

题目分析：

这道题关键要理解这个“距离”：如果是两点通过传送器之间的距离则这个距离不算，比如传送器可以把牛粪从x传送到y，而距离y-x就不能算进答案。如果是从传送器到终点的距离则计算在内。简单的来说，假如农夫1步走1个单位的距离，就是算农夫把牛粪从a运到b的最小步数；通过传送门时，由于直接传送到另一个地点，农夫要走的步数为0。

 

接下来，我们来分情况讨论：

1.农夫不通过传送门：

2.农夫通过传送点x：

3.农夫通过传送点y：

最终，我们只需要取这三种情况的最小值就可以了。

 

 

AC代码：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cmath>
 5 using namespace std;
 6
 7 int main(){
 8     int a, b, x, y;
 9     cin >> a >> b >> x >> y;
10
11     int res = 1e9;   //设为"无穷大"
12     res = min(res, abs(a-b));          //第一种情况
13     res = min(res, abs(a-x)+abs(b-y)); //第二种情况
14     res = min(res, abs(a-y)+abs(b-x)); //第三种情况
15
16     cout << res << endl;
17     return 0;
18 }