Patrick and Shopping(距离比较)
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Patrick and Shopping
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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 integersd1,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
10 20 30
60
1 1 5
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,给出两商店离家的距离,同时给出两商店间的距离,从家中出发,要经过两商店,再回到家中,求最短距离,中间可以回家。
#include<stdio.h>#include<algorithm>#include<iostream>using namespace std;int main(){int a,b,c,t;while(~scanf("%d%d%d",&a,&b,&c)){t=max(a,max(b,c));if((a+b)<=t)printf("%d\n",2*(a+b));else if((a+c)<=t)printf("%d\n",2*(a+c));else if((b+c)<=t)printf("%d\n",2*(b+c));else printf("%d\n",a+b+c);}}
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