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\n",ans);

    return 0;

}