SPOJ - Easy dijkstar problem

#include<bits/stdc++.h>
using namespace std;
#define fast()(ios_base::sync_with_stdio(false),cin.tie(NULL));
int const M= 1e9+1;
int const N= 1e4+1;
bool vis[N];
int cost[N];

vector<vector<pair<int,int> > > root;


void dijkstra(int src)
{
    int v,c;
    priority_queue<pair<int,int> >q;
    cost[src]=0;
    q.push(make_pair(0,src));
    while(!q.empty())
    {
        v=q.top().second;
        c=-q.top().first;
        q.pop();
        if(vis[v])continue;
        vis[v]=true;
        for(auto pr: root[v])
        {
            if(pr.second+c<cost[pr.first])
            {
                cost[pr.first]=pr.second+c;
                q.push(make_pair(-(pr.second+c),pr.first));
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int node, edge;
        scanf("%d %d",&node,&edge);
        root.clear();
        root.resize(node);
        int u,v,w;
        for(int i=0;i<=node;i++)cost[i]=M;
        memset(vis,false,sizeof(vis));

        for(int i=0;i<edge;i++)
        {

            scanf("%d %d %d",&u,&v,&w);
            u--,v--;
            root[u].push_back(make_pair(v,w));
        }

        scanf("%d %d",&u,&v);
        u--,v--;
        dijkstra(u);
        if(cost[v]==M)printf("NO\n");
        else printf("%d\n",cost[v]);
    }
}