A. Patrick and Shopping
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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 test(s)
input
10 20 30
output
60
input
1 1 5
output
4
Note
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.
题目大意：
有一个屋子 A 和两个超市 B 和 C，然后给定了三条道路，分别是 A--->B == d1, A---->C == d2 , B---->C == d3 的，然后让你求如何走才能
使A 到 B和C的距离最短。。
解题思路：
就是将d1 d2 d3排一下序，在跟d1 + d2 + d3比较就好了，取最小的。。。
上代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e3+5;
const int mod = 1e9+7;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
LL gcd(LL a, LL b)
{
if(b == 0)
return a;
return gcd(b, a%b);
}
int d[3];
int main()
{
while(cin>>d[0])
{
for(int i=1; i<3; i++)
cin>>d[i];
sort(d, d+3);
int sum = 0;
for(int i=0; i<3; i++)
sum += d[i];
if(sum >= 2*(d[0]+d[1]))
sum = 2*(d[0]+d[1]);
cout<<sum<<endl;
}
return 0;
}