CodeForces 17B Hierarchy

Hierarchy

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

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.

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 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

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

Examples

input

47 2 3 141 2 52 4 13 4 11 3 5

output

11

input

31 2 323 1 23 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.

因为所有的点的边都是从q大于他的点来的

所以按照q从大到小排序，对于当前点i，从j<i的里面选择一个最小的与i有边的j连接即可

#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<cmath>#include<vector>#include<string>#include<queue>#include<map>#include<algorithm>using namespace std;typedef long long LL;struct node{    int q,id;}p[1005];int a[1005][1005],d[1005],n;bool cmp(node a,node b){    return a.q>b.q;}int main(){    int i,j,k,v;    while(scanf("%d",&n)!=EOF)    {       for(i=1;i<=n;i++) for(j=1;j<=n;j++) a[i][j]=-1;       for(i=1;i<=n;i++)       {           scanf("%d",&p[i].q);           p[i].id=i;       }       sort(p+1,p+1+n,cmp);       for(i=1;i<=n;i++)       {           d[p[i].id]=i;       }       scanf("%d",&k);       while(k--)       {           scanf("%d%d%d",&i,&j,&v);           i=d[i],j=d[j];           if(a[i][j]==-1) a[i][j]=v;           else a[i][j]=min(a[i][j],v);       }       int flag=0,ans=0;       for(i=2;i<=n;i++)       {           v=10000000;           for(j=1;j<i;j++)           {               if(p[j].q>p[i].q&&a[j][i]!=-1)                v=min(v,a[j][i]);           }           if(v==10000000) flag=1;           ans+=v;           if(flag) break;       }       if(flag) ans=-1;       printf("%d\n",ans);    }    return 0;}