Patrick and Shopping

I - Patrick and Shopping

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

SubmitStatus

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 ad₁ meter long road between his house and the first shop and ad₂ meter long road between his house and the second shop. Also, there is a road of lengthd₃ 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.

题意：你从家出发去两个商店买东西，你家和两个商店之间，以及两个商店之间分别有一条路，现在要求你走的最短的路程

思路：你一共有四种选择：d1+d2+d3，d1*2+d2*2，d1*2+2*d3，d2*2+d3*2；求出其中最小的就可以了。

#include<stdio.h>#include<algorithm>using namespace std;int main(){    int d1,d2,d3;    while(~scanf("%d%d%d",&d1,&d2,&d3))    {        int ans=d1+d2+d3;        ans=min(ans,d1*2+d2*2);        ans = min (ans,d1*2+2*d3);        ans = min(ans,d2*2+d3*2);        printf("%d\n",ans);    }    return 0;}