Patrick and Shopping

最新推荐文章于 2020-02-09 11:15:11 发布

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最新推荐文章于 2020-02-09 11:15:11 发布

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分类专栏： OJ练习

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本题描述了Patrick为了迎接朋友Spongebob的到访，需要前往两个商店购买物品，并返回家的过程。目标是最小化总行走距离。输入包含三个整数，分别代表从家到第一个商店、家到第二个商店及两商店之间的道路长度。程序通过比较不同的路线组合来计算最小行走距离。

C - Patrick and Shopping

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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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.

$\text{[math]}$

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

Input

1 1 5

Output

Hint

The first sample is shown on the picture in the problem statement. One of the optimal routes is: house $\text{[math]}$ first shop $\text{[math]}$ second shop $\text{[math]}$ house.

In the second sample one of the optimal routes is: house $\text{[math]}$ first shop $\text{[math]}$ house $\text{[math]}$ second shop $\text{[math]}$ house.

AC代码：

#include<stdio.h>
#include<math.h>
__int64 min(__int64 x,__int64 y)
{
	if (x>=y)
	return y;
	else
	return x;
}
int main()
{
	__int64 d1,d2,d3;
	while(scanf("%I64d%I64d%I64d",&d1,&d2,&d3)!=EOF)
	{
		printf("%I64d\n",min(min(d1+d2+d3,2*d1+2*d2),min(2*d2+2*d3,2*d3+2*d1)));
	}
	return 0;
}