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: ?employee ai 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.
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 qj (0?≤?qj?≤?106)― the employees' qualifications. The following line contains number m (0?≤?m?≤?10000) ― amount of received applications. The following mlines 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
Output the only line ― the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Sample test(s) 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
Note
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.
预处理+贪心就可以了，因为根只有一个，所以在输入的时候预处理子节点的最小值贪心就好了=。=
#include
#include
#include
#include
#include
using namespace std; const int INF=0x3ffffff; int f[1100]; int main() { int n,m; int temp,u,v,w; while(cin>>n) { for(int i=1;i<=n;i++) { cin>>temp; f[i]=INF; } cin>>m; for(int i=1;i<=m;i++) { cin>>u>>v>>w; if(f[v]>w)//预处理最小值 f[v]=w; } int flag=1,k=1; int ans=0; for(int i=1;i<=n;i++)//只能有一个INF，即根节点 { if(f[i]==INF&&k) { ans-=INF; k=0; } else if(f[i]==INF) { flag=0; break; } ans+=f[i]; } if(flag) cout<