(POJ - 1847) tram (shortest circuit)

Topic link ：1847 -- Tram

The question ： Yes n A little bit , Then these points and other points have some paths , Each point is a switch , The switch can only go one way in one direction , The first number is the default switch pointing to , No rotation . It's the default pointing. In fact, you only need to rotate 0 Time , Other paths only need to rotate 1 Time , Whatever it is , just 1 Time . seek a To b Minimum number of rotations

This problem is a bare template with the shortest path , But the length of the edge he gave was only 0 and 1, Is that the distance of the first point given is 0, The distance between other points is 1, This place has stuck me for a long time , At first, I thought that the current point was the number and the given distance was the number -1, In fact, as long as it is not the first one to give , Then the distance is 1, This place needs attention , The rest of the set of the shortest path template will pass , Put the code below ：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
const int N=1003;
int h[N],w[N*N],ne[N*N],e[N*N],idx,d[N];
bool vis[N];
int n,b,en;
typedef pair<int,int>PII;
void add(int x,int y,int z)
{
	e[idx]=y;
	w[idx]=z;
	ne[idx]=h[x];
	h[x]=idx++;
}
void dijkstra(int x)
{
	memset(d,0x3f,sizeof d);
	priority_queue<PII,vector<PII>,greater<PII> >q;
	d[x]=0;
	q.push({d[x],x});
	while(!q.empty())
	{
		int begin=q.top().second;
		q.pop();
		if(vis[begin]) continue;
		vis[begin]=true;
		for(int i=h[begin];i!=-1;i=ne[i])
		{
			int j=e[i];
			if(d[j]>d[begin]+w[i])
			{
				d[j]=d[begin]+w[i];
				q.push({d[j],j});
			}
		}
	}
}
int main()
{
	memset(h,-1,sizeof h);
	cin>>n>>b>>en;
	for(int i=1;i<=n;i++)
	{
		int cnt,t;
		scanf("%d",&cnt);
		for(int j=1;j<=cnt;j++)
		{
			scanf("%d",&t);
			if(j==1) add(i,t,0);
			else add(i,t,1);
		}
	}
	dijkstra(b);
	if(d[en]==0x3f3f3f3f) puts("-1");
	else printf("%d",d[en]);
	return 0;
}