luogu P1346 电车 最短路

显然结论

一个点不会经过了两次，一个阀门不会反复开关。第一条边为0，其余边为1。

#include <cstdio>
#include <queue>
using namespace std;
int head[120],dis[120],to[30000],nxt[30000],val[30000];
bool vis[120];
int cnt,s,t,n;

struct pot
{
    int x,dis;
    pot (int _x = 0,int _dis = 0) : x(_x),dis(_dis) {}
    friend bool operator < (pot a,pot b)
    {
        return a.dis > b.dis;
    }
};
priority_queue <pot> que;
void add(int x,int y,int v)
{
    nxt[++cnt] = head[x];
    to[cnt] = y;
    head[x] = cnt;
    val[cnt] = v;
}
void dijkstra()
{
    for(int i=0; i<= n; i++) dis[i]=2e9;
    que.push(pot(s,0));
    dis[s]=0;
    while(!que.empty())
    {
        pot now=que.top();
        que.pop();
        if(vis[now.x]) continue;
        vis[now.x]=true;
        for(int i = head[now.x]; i; i=nxt[i])
        {
            if(dis[to[i]]>dis[now.x]+val[i])
            {
                dis[to[i]]=dis[now.x]+val[i];
                que.push(pot(to[i],dis[to[i]]));
            }
        }
    }
}
int main()
{
    scanf("%d%d%d",&n,&s,&t);
    int tk,tmp;
    for (int i = 1;i <= n;i++)
    {
        scanf("%d",&tk);
        if (tk)
        {
            scanf("%d",&tmp); 
            add(i,tmp,0);
        }
        for (int o = 2;o <= tk;o++)
        {
            scanf("%d",&tmp);
            add(i,tmp,1);
        }
    }
    dijkstra();
    if (dis[t] == dis[0]) printf("-1
");
    else printf("%d
",dis[t]);
    return 0;
}