“浪潮杯”第九届山东省ACM大学生程序设计竞赛重现赛 C-Cities

Posted 2020-11-06 霜雪千年

题目描述：There are n cities in Byteland, and the i_th city has a value a_i. The cost of building a bidirectional road between two cities is the sum of their values. Please  calculate the minimum cost of connecting these cities, which means any two cities can reach each other.输入描述：第一行是一个表示测试用例数量的整数  t(t<=10⁵)。对于每个测试用例，第一行是一个整数n(n<=10⁵)，代表城市的数量，第二行是n个 正整数ai(a_i<=10⁵) ，代表它们的值。

输出描述

For each test case, output an integer, which is the minimum cost of connecting these cities.

示例1

输入

2
4
1 2 3 4
1
1

输出

12
0
解题思路：水题！将最低成本的城市分别与n-1个城市直接相连，这就是修道路所花费最低成本的方法。
AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+5;
 4 int a[maxn];
 5 int main()
 6 {
 7     int t,n;long long sum;//这里要用long long
 8     cin>>t;
 9     while(t--){
10         cin>>n;sum=0;
11         for(int i=0;i<n;++i)
12             cin>>a[i];
13         sort(a,a+n);
14         for(int i=1;i<n;++i)
15             sum+=a[i];
16         if(sum>0)cout<<(sum+a[0]*(n-1))<<endl;
17         else cout<<‘0‘<<endl;
18     }
19     return 0;
20 }