傳送門

方法一：分層圖

#include<bits/stdc++.h>
#define per(i,a,b) for(int i=a;i<=b;i++)
#define mod 1000000007
using namespace std;
typedef long long ll;
const ll inf =23333333333333333LL;
const double eps=1e-8;
int T;
int read(){
    char ch=getchar();
    int res=0,f=0;
    while(ch<‘0‘ || ch>‘
 
9‘){f=(ch==‘-‘?-1:1);ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){res=res*10+(ch-‘0‘);ch=getchar();}
    return res*f;
}
// ------------------------head
const int siz=100005;
map<string,int> mp;
int mcnt=0,n,m;
int d;
char x[15],y[15],schar[15],echar[15];
int head[siz],Enum=0;
ll dis[siz][
 
2];
bool vis[siz][2];
int st,ed;
struct Edge{int to,w,ne;}edge[siz*5];
void add_edge(int a,int b,int c){
    edge[Enum].to=b;
    edge[Enum].w=c;
    edge[Enum].ne=head[a];
    head[a]=Enum++;
}
void init()
{
    memset(head,-1,sizeof(head));
    Enum=0;
    mp.clear();
    mcnt=0;
} 
 
int _hash(char *s){//char*作為實參傳給string形參會自動變
    if(mp.count(s)>0)return mp[s];//string作為形參傳給char*要轉變s.c_str();
    else return mp[s]=mcnt++;
}
struct Qnode{
    int u;
    ll _dis;
    int layer;//位於分層圖的哪層
    Qnode(){}
    Qnode(ll a,int b,int c):_dis(a),u(b),layer(c){}
    bool operator<(const Qnode&rhs)const{return _dis>rhs._dis;}//一定要加const
};
void Dijkstra(){
    priority_queue<Qnode>que;
    while(!que.empty())que.pop();
    memset(vis,false,sizeof(vis));
    per(i,0,n){dis[i][0]=inf;dis[i][1]=inf;}
    que.push(Qnode(0LL,st,0));
    dis[st][0]=0; //註意初始化的是st點，而不是0點，這裏沒註意到！！！
    vis[st][0]=true;
    Qnode tmp;
    while(!que.empty()){
        tmp=que.top();que.pop();
        int u=tmp.u,layer=tmp.layer;
        ll _dis=tmp._dis;
        //printf("dis:%lld   u:%d   layer:%d\n",_dis,u,layer);//
        vis[u][layer]=true;
        //if(u==ed && layer==1)return;//
        //printf("dis[%d]:%lld\n",u,dis[u][0]);//
        for(int i=head[u];i!=-1;i=edge[i].ne){
            int v=edge[i].to,w=edge[i].w;
            //printf("w:%d\n",w);exit(0);//
            //printf("v:%d  \n");//
            //printf("dis[u][layer]:%lld\n",dis[u][layer]);//
            //printf("dis[v][layer+1]:%lld  dis[v][layer]:%lld  w:%d\n",dis[v][layer+1],dis[v][layer],w);//
            //if(dis[v][layer+1]>dis[u][layer]+w/2)printf("cas1 ok\n");else printf("cas1 no\n");//
            if(layer<1 && !vis[v][layer+1] &&dis[v][layer+1]>dis[u][layer]+w/2){
                dis[v][layer+1]=dis[u][layer]+w/2;
                //printf("Push  v%d  dis:%lld  layer:%d\n",v,dis[v][layer],layer+1);//
                que.push(Qnode(dis[v][layer+1],v,layer+1));
            }
            if(!vis[v][layer] && dis[v][layer]>dis[u][layer]+w){
                dis[v][layer]=dis[u][layer]+w;
                que.push(Qnode(dis[v][layer],v,layer));
            }
        }
    }
}

int main()
{
    while(scanf("%d %d",&n,&m)!=EOF){
        init();
        per(i,1,m){
            scanf("%s %s %d",x,y,&d);
            int xn=_hash(x),yn=_hash(y);
            //printf("%s:%d %s:%d d:%d\n",x,xn,y,yn,d);//
            add_edge(xn,yn,d);
        }
        scanf("%s %s",schar,echar);
        if(mp.count(schar)==0||mp.count(echar)==0){printf("-1\n");continue;}
        st=_hash(schar);ed=_hash(echar);
        //printf("%s:%d  %s:%d\n",schar,st,echar,ed);//
        Dijkstra();
        ll ans=min(dis[ed][0],dis[ed][1]);
        if(ans==inf)printf("-1\n");
        else printf("%lld\n",ans);
    }

    return 0;
}

方法二：反向建圖

hdu3499(分層圖最短路 or 反向建圖)