題目大意：
給你n個點，m條邊，讓你最多可以使k條邊的權值為0，問你1到n的最短路。
分析：
dijsktra的距離陣列可以多開一維，記錄已經讓幾條邊的權值為0，即分層圖最短路

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
#include <set>
#include <vector>
using namespace std;
const int maxn = 100000 + 10
 
;
typedef long long ll;
struct HeapNode
{
    ll d;
    int u,num;
    bool operator<(const HeapNode &rhs)const
    {
      return d>rhs.d;
    }
};
struct Edge
{
    int fro,to;
    ll cost;
    int id;
};
const ll INF = 1e18;
struct Dij
{
    vector<ll>G[maxn];
    vector<Edge>
 
edges;
    ll d[maxn][15];
    bool done[maxn][15];
    int n,m,k_;
    void init(int n,int k)
    {
        this->n=n;
        this->k_=k;
        for(int i=0;i<=n;i++) G[i].clear();
        edges.clear();
    }
    void addedge(int from,int to,int cost,int id)
    {
        edges.push_back
 
((Edge){from,to,cost,id});
        m=edges.size();
        G[from].push_back(m-1);
        //cout<<G[1].size()<<endl;
    }
    void dij(int s)
    {
        for(int i=0;i<=n;i++)
            for(int j=0;j<=10;j++) d[i][j]=INF;
        memset(done,0,sizeof(done));
        priority_queue<HeapNode>Q;
        d[s][0]=0;
        Q.push((HeapNode){0,s,0});
        while(!Q.empty())
        {
            HeapNode now=Q.top(); Q.pop();
            int u=now.u;
            int k=now.num;
            if(done[u][k]) continue;
            done[u][k]=true;
           // cout<<u<<" "<<G[u].size()<<endl;
            for(int i=0;i<G[u].size();i++)
            {
                Edge &e=edges[G[u][i]];
                //cout<<d[e.to][k]<<" "<<d[u][k]<<" "<<e.cost<<endl;
                if(d[e.to][k]>d[u][k]+e.cost)
                {

                    d[e.to][k]=d[u][k]+e.cost;
                    //cout<<e.to<<" "<<u<<" "<<k<<" "<<d[e.to][k]<<endl;
                    Q.push((HeapNode){d[e.to][k],e.to,k});
                }
                if(k==k_) continue;
                if(d[e.to][k+1]>d[u][k])
                {

                    d[e.to][k+1]=d[u][k];
                   // cout<<e.to<<" "<<u<<" "<<k+1<<" ..."<<d[e.to][k+1]<<endl;
                    Q.push((HeapNode){d[e.to][k+1],e.to,k+1});
                }

            }
        }
    }
    void print(int t)
    {
        ll ans=INF;
        for(int i=0;i<=k_;i++)
        {
            ans=min(ans,d[t][i]);
        }
        printf("%lld\n",ans);
    }
}D;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        D.init(n,k);
        for(int i=0;i<m;i++)
        {
            int u,v;
            ll c;
            scanf("%d%d%lld",&u,&v,&c);
            D.addedge(u,v,c,i);
        }
        D.dij(1);
        D.print(n);
    }
    return 0;
}