典型的分層圖的題，因為k的取值範圍很小，所以我們需要另外再建立k層圖，其實就是相當於跨層建邊，主要就是這個過程怎麼實現，剩下的就是一個裸的dij了。看到一份用二維陣列存的程式碼，感覺要比我的程式碼寫的簡單，過兩天敲一下試試，傳送門

AC程式碼：

#include <bits/stdc++.h>
#define maxn 1200100
#define ll long long
using namespace std;
struct node{
  int next;
  int to;
  int val;
}Edge[maxn << 1];
struct Node{
  int val,cnt;
  bool operator < (const Node &a) const {
    if(a.val == val) return a.cnt < cnt;
    return a.val < val;
  }
}Next,Now;
int n,m,k,s,e;
int head[maxn],num,dist[maxn];
bool vis[maxn];
priority_queue<Node> q;

void init(){
  memset(head,-1,sizeof(head));
  num = 0;
}

void add(int u,int v,int val){
  Edge[num].to = v;
  Edge[num].val = val;
  Edge[num].next = head[u];
  head[u] = num++;
}

void dijkstra(){
  memset(vis,false,sizeof(vis));
  memset(dist,0x3f,sizeof(dist));
  Now.val = 0;
  Now.cnt = s;
  dist[s] = 0;
  q.push(Now);
  while(!q.empty()){
    Now = q.top();
    q.pop();
    int flag = Now.cnt;
    if(vis[flag])continue;
    vis[flag] = true;
    for(int i=head[flag]; i != -1; i = Edge[i].next){
      int u = Edge[i].to;
			if(!vis[u] && dist[u] > dist[flag] + Edge[i].val){
				dist[u] = dist[flag] + Edge[i].val;
				Next.cnt = u;
				Next.val = dist[u];
				q.push(Next);
			}
    }
  }
  int ans = 0x3f3f3f3f;
  for(int i=0;i<=k;i++){
    ans = min(ans, dist[e + i * n]);
  }
  printf("%d\n",ans);
}

int main()
{
  init();
  scanf("%d%d%d",&n,&m,&k);
  scanf("%d%d",&s,&e);
  s++;e++;
  for(int i=0;i<m;i++){
    int u,v,w;
    scanf("%d%d%d",&u,&v,&w);
    u++;v++;
    for(int j=0;j<=k;j++){
      add(u+j*n, v+j*n, w);
      add(v+j*n, u+j*n, w);
      if(j != k){
        add(u+j*n, v+(j+1)*n, 0);
        add(v+j*n, u+(j+1)*n, 0);
      }
    }
  }
  dijkstra();
  return 0;
}