A*K短路模板，详见https://blog.csdn.net/z_mendez/article/details/47057461
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int N=10005;
const long long inf=1e15;
int n,m,k,s,t,h[N],cnt,x[N],y[N],z[N],c[N];
long long dis[N],ans[N];
bool v[N];
struct qwe
{
int ne,to,va;
}e[N];
struct dian
{
int x;
long long g,h;//g是已走过的路，h是到目的地的最短路
dian(int X=0,long long G=0,long long H=0)
{
x=X,g=G,h=H;
}
bool operator < (const dian &a) const
{
return g+h>a.g+a.h;//priority是大根堆，把小于重载成大于相当于改成小根堆
}
};
int read()
{
int r=0,f=1;
char p=getchar();
while(p>'9'||p<'0')
{
if(p=='-')
f=-1;
p=getchar();
}
while(p>='0'&&p<='9')
{
r=r*10+p-48;
p=getchar();
}
return r*f;
}
void add(int u,int v,int w)
{
cnt++;
e[cnt].ne=h[u];
e[cnt].to=v;
e[cnt].va=w;
h[u]=cnt;
}
void aster()
{
priority_queue<dian>q;
for(int i=1;i<=k;i++)
ans[i]=-1;
q.push(dian(n,0,dis[n]));
int tot=0;
while(!q.empty())
{
int u=q.top().x;
long long g=q.top().g;
q.pop();
c[u]++;
if(c[u]>k)
continue;//如果当前点第k次出队,则当前点已经不能给前k短路贡献了
if(u==1)
ans[++tot]=g;
if(c[1]==k)
return;
for(int i=h[u];i;i=e[i].ne)
q.push(dian(e[i].to,g+e[i].va,dis[e[i].to]));
}
}
int main()
{
n=read(),m=read(),k=read();
for(int i=1;i<=m;i++)
{
x[i]=read(),y[i]=read(),z[i]=read();
if(x[i]<y[i])
swap(x[i],y[i]);
add(y[i],x[i],z[i]);
}
queue<int>q;
for(int i=1;i<=n;i++)
dis[i]=inf;
dis[1]=0,v[1]=1,q.push(1);
while(!q.empty())
{
int u=q.front();
q.pop();
v[u]=0;
for(int i=h[u];i;i=e[i].ne)
if(dis[e[i].to]>dis[u]+e[i].va)
{
dis[e[i].to]=dis[u]+e[i].va;
if(!v[e[i].to])
v[e[i].to]=1,q.push(e[i].to);
}
}
memset(h,0,sizeof(h));
cnt=0;
for(int i=1;i<=m;i++)
add(x[i],y[i],z[i]);
aster();
for(int i=1;i<=k;i++)
printf("%lld\n",ans[i]);
return 0;
}