Cities

本文介绍了一道关于计算连接多个城市使得任意两城市间可达的最小成本的算法题。问题中,每个城市的值代表其特性,双向道路的成本为两城市值之和。文章通过示例解释了如何使用贪心算法求解此问题,并提供了完整的C++代码实现。

问题 C: Cities

时间限制: 1 Sec  内存限制: 128 MB
提交: 87  解决: 61
[提交][状态][讨论版][命题人:admin]

题目描述

There are n cities in Byteland, and the ith city has a value ai. 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

The first line is an integer T(T≤10^5), representing the number of test cases.
For each test case, the first line is an integer n(n≤10^5), representing the number of cities, the second line are n positive integers ai(ai≤10^5), representing their values.
 

输出

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

样例输入

2
4
1 2 3 4
1
1

样例输出

12
0


#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        if(n==1)
        {
            cout<<0<<endl;
            continue;
        }
        int a[100005];
        cin>>a[0];
        int minx = a[0];
        int index = 0;
        for(int i=1;i<n;i++)
        {
            cin>>a[i];
            if(a[i]<minx)
            {
                minx = a[i];
                index = i;
            }
        }
        long long int ans = 0;
        for(int i=0;i<n;i++)
        {
            if(i!=index)
            {
                ans+=minx+a[i];
            }
        }
        cout<<ans<<endl;
    }
}

贪心题

转载于:https://www.cnblogs.com/hao-tian/p/9080582.html

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