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.

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

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：

2019 GDUT Rating Contest II : Problem F. Teleportation
题面：
题目描述：
题目分析：

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 }

2019 GDUT Rating Contest II : Problem F. Teleportation 题面： 题目描述： 题目分析：

题目描述：

题目分析：

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2019 GDUT Rating Contest II : Problem F. Teleportation 题面：题目描述：题目分析：