codeforce 17 B

http://vjudge.net/contest/view.action?cid=48014#problem/B

Description

Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employeea_i is ready to become a supervisor of employee b_i at extra cost c_i». The qualification q_j of each employee is known, and for each application the following is true: q_ai > q_bi.

Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.

Input

The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers q_j (0 ≤ q_j ≤ 10⁶)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: a_i, b_i and c_i (1 ≤ a_i, b_i ≤ n, 0 ≤ c_i ≤ 10⁶). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application q_ai > q_bi.

Output

Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.

Sample Input

Input

4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5

Output

11

Input

3
1 2 3
2
3 1 2
3 1 3

Output

-1

Hint

In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.

题目大意：有n个人在一家公司，公司在这几个人中建立一个树形关系，使得每一个人都在其中，一个人是另一个人上司（父节点），每一个中上下属关系会有一定的费用值，求怎样安排才能使得费用最低。

解题思路：本来是想着用并查集做来，可是对于题目的要求，会出现不是树的情况，那样不好处理。后来看看学长以前的代码发现我其实是想的复杂了==

                    首先对每一个员工的地位值进行从小到大排序，那么地位最高的那个人是不会有上级的其余的人必须都要满足有上级才 行，否则输出-1，在满足每一个人都有上级                       的情况下，选择二者搭档费用最小的搭档，最后的费用总值即为所求。

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

struct per
{
    int num,q;
};
int cmp(per a,per b)
{
    return a.q<b.q;
}
int main()
{
    int n,m;per data[1005];int ap[10005][3];
    while(cin>>n)
    {
        for(int i=0;i<n;i++)
        cin>>data[i].q,data[i].num=i+1;
        sort(data,data+n,cmp);
        cin>>m;
        for(int i=0;i<m;i++)
        cin>>ap[i][0]>>ap[i][1]>>ap[i][2];
        int f=1,ans=0;
        for(int i=0;i<n-1;i++)
        {
            int w=99999999;int flag=0;
            for(int j=0;j<m;j++)
            {
                if(data[i].num==ap[j][1]&&ap[j][2]<w)
                w=ap[j][2],flag=1;
            }
            if(flag==0)
            {
                f=0;break;
            }
            ans+=w;
        }
        if(!f)
        cout<<-1<<endl;
        else
        cout<<ans<<endl;
    }
    return 0;
}