17B. Hierarchy

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: «employeeai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.

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

The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 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: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 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 qai > qbi.

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

47 2 3 141 2 52 4 13 4 11 3 5output
11input
31 2 323 1 23 1 3output
-1Note
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.
借鉴大神的做法
#include<cstdio>#include<algorithm>using namespace std;const int maxn=0x3f3f3f3f;int result(int n,int x[]){	int sum=0,kase=1;	for(int i=1;i<=n;++i)	{		if(x[i]==maxn)		{			sum-=maxn;			if(!kase)			{				return -1;			}			kase=0;		}		sum+=x[i];	}	return sum;}int main(){	int n;	while(~scanf("%d",&n))	{		int temp,x[1005]={0};		for(int i=1;i<=n;++i)		{			scanf("%d",&temp);			x[i]=maxn;		}				int m,a,b,c;		scanf("%d",&m);		for(int i=0;i<m;++i)		{			scanf("%d%d%d",&a,&b,&c);			if(x[b]>c)//留下最少花费的 			{				x[b]=c;			}		}		PRintf("%d/n",result(n,x));	}	return 0;}