題目描述

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

輸入描述：

The first line is an integer T（T<= 1e5） representing the number of test cases.

For each test case, the first line is an integer n(n <= 1e5), representing the number of cities, the second line are n positive integers a (a <= 1e5), representing their values.

輸出描述：

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

示例1

輸入

2

4

1 2 3 4

1

1

輸出

12

0

/*題意: Byteland有n個城市，i城市有價值a。 在兩個城市之間建立雙向道路的成本是它們的價值的總和。 請計算連線這些城市的最低成本，這意味著任何兩個城市都可以相互聯絡*/

思路:

第一眼以為最小生成樹，看了點數10的5次方，需要求出每個邊的值，邊樹大約有10的9次方個，然後prim演算法進行尋找，資料量太大，定會超時。。。。。

由此想到貪心演算法：

想要將任意兩個城市連線且道路花費之和最小，那可使每條道路的花費最少，道路的花費等於兩端城市value之和，由此可知，只要每個城市與最小value的那個城市相連通，所得的花費必定是最小的。

因此，將最小value的城市放於中間與其他城市一一相連。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define p 100001
using namespace std;
int main(){
    int t;
    int a[p];
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
         if(n==1){
            printf("0\n");
            continue;
        }
        sort(a,a+n);
        int minn=a[0];
        long long sum=0;
        for(int i=1;i<n;i++){
            sum=sum+a[i]+minn;
        }
        printf("%lld\n",sum);
    }
}