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社区首页 >专栏 >【CodeForces 599A】D - 特别水的题4- Patrick and Shopping

【CodeForces 599A】D - 特别水的题4- Patrick and Shopping

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饶文津
发布2020-05-31 23:29:40
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发布2020-05-31 23:29:40
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文章被收录于专栏:饶文津的专栏饶文津的专栏

Description

Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a d1 meter long road between his house and the first shop and a d2 meter long road between his house and the second shop. Also, there is a road of length d3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.

Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.

Input

The first line of the input contains three integers d1, d2, d3 (1 ≤ d1, d2, d3 ≤ 108) — the lengths of the paths.

  • d1 is the length of the path connecting Patrick's house and the first shop;
  • d2 is the length of the path connecting Patrick's house and the second shop;
  • d3 is the length of the path connecting both shops.

Output

Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.

Sample Input

Input

10 20 30

Output

60

Input

1 1 5

Output

4

Hint

The first sample is shown on the picture in the problem statement. One of the optimal routes is: house - first shop - second shop -house.

In the second sample one of the optimal routes is: house - first shop - house - second shop - house.

这题只要分4种情况,选择最短的路径就可以了

代码语言:javascript
复制
#include<stdio.h>
#include<algorithm>
using namespace std;
long long d1,d2,d3,ans;
int main(){

    scanf("%lld%lld%lld",&d1,&d2,&d3);
    ans=min(min(d1*2+d3*2,d2*2+d3*2),min(d1+d2+d3,d1*2+d2*2));
    printf("%d",ans);
    return 0;
}
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原始发表:2015-12-14 ,如有侵权请联系 cloudcommunity@tencent.com 删除

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