题解
最短路模型，这很显然。
顺带一提，不要用SPFA，真心卡成狗。1
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using namespace std;
typedef long long ll;
int read(){
int f=1,x=0;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-f;c=getchar();}
while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
return f*x;
}
struct Edge{
int to,_next,cost;
};
struct Node{
int h,e,dis;
Node(int h_,int e_,int d_){
h=h_,e=e_,dis=d_;
}
Node(){}
};
bool operator <(const Node &n1,const Node &n2){
return n1.dis>n2.dis;
}
priority_queue<Node> pq;
Edge edge[100005];
int n,m,k,at[10005]={0},F[10005][21]={0},cnt=0;
void addedge(int u,int v,int c){
edge[++cnt].to=v,
edge[cnt].cost=c,
edge[cnt]._next=at[u],
at[u]=cnt;
}
void Dij(){
F[1][0]=0,pq.push(Node(1,0,0));
Node tmp;
int h_,e_,d_,v_,c_;
while(!pq.empty()){
tmp=pq.top(),pq.pop();
if(tmp.dis>F[tmp.h][tmp.e])continue;
h_=tmp.h,e_=tmp.e,d_=tmp.dis;
for(int i=at[h_];i;i=edge[i]._next){
v_=edge[i].to,c_=edge[i].cost;
if(F[v_][e_]>F[h_][e_]+c_)
F[v_][e_]=F[h_][e_]+c_,
pq.push(Node(v_,e_,F[v_][e_]));
if(e_<k&&F[v_][e_+1]>F[h_][e_])
F[v_][e_+1]=F[h_][e_],
pq.push(Node(v_,e_+1,F[v_][e_+1]));
}
}
}
void init(){
n=read(),m=read(),k=read();
int u,v,c;
for(int i=1;i<=m;i++){
u=read(),v=read(),c=read();
addedge(u,v,c),addedge(v,u,c);
}
}
void solve(){
memset(F,0x3F,sizeof(F));
Dij();
int ans=INF*2;
for(int i=0;i<=k;i++)ans=min(ans,F[n][i]);
printf("%d\n",ans);
}
int main(){
init();
solve();
return 0;
}