易错点：

• 将边权转化为0/1判定性问题.

#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;
const int MAXN=1000010,MAXM=1000010,INF=2100000000;
int n;
struct Edge{
    int from,to,w,nxt;
}e[MAXM];
int head[MAXN],edgeCnt=0;
void addEdge(int u,int v,int w){
    e[++edgeCnt].from=u;
    e[edgeCnt].to=v;
    e[edgeCnt].w=w;
    e[edgeCnt].nxt=head[u];
    head[u]=edgeCnt;
}
int dis[MAXN],s;
struct Node{
    int nowPoint,nowValue;
    bool operator <(const Node &a)const{
        return a.nowValue<nowValue;
    }
};
int nowK=-1;//现在二分的值 
int dijkstra(){
    priority_queue<Node> q;
    for(int i=1;i<=n;i++){
        dis[i]=INF;
    }
    q.push(Node{s,0});dis[s]=0;
    while(!q.empty()){
        Node nowNode=q.top();q.pop();
        int nowPoint=nowNode.nowPoint,nowValue=nowNode.nowValue;
        if(dis[nowPoint]!=nowValue)continue;
        for(int i=head[nowPoint];i;i=e[i].nxt){
            int nowV=e[i].to;
            int tmpW=e[i].w>nowK?1:0;
            if(dis[nowV]>dis[nowPoint]+tmpW){
                dis[nowV]=dis[nowPoint]+tmpW;
                q.push(Node{nowV,dis[nowPoint]+tmpW});
            }
        }
    }
    return dis[n];
}
const int MAXL=1000005;
int main(){
    int m,k;//边数和免费量 
    scanf("%d%d%d",&n,&m,&k);
    s=1;
    for(int i=1;i<=m;i++){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        addEdge(u,v,w);
        addEdge(v,u,w);
    }
    dijkstra();
    int l=0,r=MAXL;
    int ans=2000000001;
    while(l<r){
        int mid=(l+r)>>1;
        nowK=mid;
        if(dijkstra()>k)l=mid+1;
        else r=mid;
    }
    if(l==MAXL){
        printf("-1\n");
    }else printf("%d\n",r);
    return 0;
}