k<=10，所以每用一次机会就跳到一个新图中，
每一个图按原图建边，相邻两图中建边权为0的边
补一补dj，好像我以前觉得dj特别难，hhhhh
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<iostream>
#define N 500500
using namespace std;
int n,m,k,S,T;
struct point{
int st,dis;
bool operator < (const point &a)const{
return dis>a.dis;
}
}p[N];
int e=1,head[N],dis[N];
bool bo[N];
struct edge{
int u,v,w,next;
}ed[5000500];
void add(int u,int v,int w){
ed[e].u=u; ed[e].v=v; ed[e].w=w;
ed[e].next=head[u]; head[u]=e++;
}
priority_queue<point> q;
int dijkstra(){
memset(dis,0x7f,sizeof dis);
memset(bo,0,sizeof bo);
dis[S]=0;q.push((point){S,0});
while(!q.empty()){
point now=q.top();q.pop();
if(bo[now.st]) continue;
bo[now.st]=1;
for(int i=head[now.st];i;i=ed[i].next)
if(dis[now.st]+ed[i].w<dis[ed[i].v]){
dis[ed[i].v]=dis[now.st]+ed[i].w;
q.push((point){ed[i].v,dis[ed[i].v]});
}
}
int ans=0x7fffffff;
for(int i=0;i<=k;i++)
ans=min(ans,dis[i*n+T]);
return ans;
}
int main(){
scanf("%d%d%d",&n,&m,&k);
scanf("%d%d",&S,&T);S++;T++;
int u,v,w;
for(int i=1;i<=m;i++){
scanf("%d%d%d",&u,&v,&w);
u++;v++;
for(int j=0;j<=k;j++){
add(j*n+u,j*n+v,w),add(j*n+v,j*n+u,w);
if(j<k)add(j*n+u,(j+1)*n+v,0),add(j*n+v,(j+1)*n+u,0);
}
}
int ans=dijkstra();
printf("%d
",ans);
return 0;
}