把费用限制放在最短路数组判断前后有什么影响？

#include<bits/stdc++.h>
using namespace std;
int n,m,t,l,r,b,c[10007],ver[100007],edge[100007],head[100007],nextt[100007];
bool flag[100007];
long long d[100007];
priority_queue <pair<int,int> > q;

bool dijkstra(int k)
{
    if(k<c[1]||k<c[n])return 0;
    memset(flag,0,sizeof(flag));
    for(int i=1;i<=n;i++)
        d[i]=0x7fffffff;
    q.push(make_pair(0,1));
    d[1]=0;
//ac代码    
    for(int i=1;i<=n;i++)
        if(c[i]>k)flag[i]=1;
    while(!q.empty())
    {
        int x=q.top().second;
        q.pop();
        if(flag[x])continue;
        flag[x]=1;
        for(int i=head[x];i;i=nextt[i])
        {
            int y=ver[i],z=edge[i];
            if(d[y]>d[x]+z)
            {
                d[y]=d[x]+z;
                q.push(make_pair(-d[y],y));
            }
        }
    }
   //ac代码结束

   /*72分
    while(!q.empty())
    {
        int x=q.top().second;
        q.pop();
        if(flag[x])continue;
        flag[x]=1;
        for(int i=head[x];i;i=nextt[i])
        {
            int y=ver[i],z=edge[i];
            if(d[y]>d[x]+z&&c[i]<=k)
            {
                d[y]=d[x]+z;
            q.push(make_pair(-d[y],y));
            }
        }

    }
    */

    if(d[n]>=b)return 0;
    else return 1;

}

int main()
{
    scanf("%d%d%d",&n,&m,&b);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&c[i]);
        r=max(r,c[i]);
    }
    l=max(c[1],c[n]);
    for(int i=1;i<=m;i++)
    {
        int x,y,z;
        cin>>x>>y>>z;
        ver[++t]=y;
        edge[t]=z;
        nextt[t]=head[x];
        head[x]=t;
        ver[++t]=x;
        edge[t]=z;
        nextt[t]=head[y];
        head[y]=t;
    }
    if(!dijkstra(r))
    {
        cout<<"AFK"<<endl;
        return 0;
    } 
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(dijkstra(mid))r=mid-1;
        else l=mid+1;
    }
    printf("%d\n",l);
    return 0;
 }