所以要枚举最短路上的等级范围从[level[1]-m,level[1]]到[level[1],level[1]+m] ！！ 
这其实还是符合逻辑的，比如酋长的曾祖父之类的咯~

附上SPFA AC代码 1164K 16MS：
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <cmath>
#include <iomanip>
#include <algorithm>
using namespace std;
const int maxn=200;
const int INF=999999999;
struct Edge
{
	int to,cost;
	Edge *next;
	Edge():to(0),cost(0),next(0){}
	Edge(int TO,int COST,Edge *NEXT):to(TO),cost(COST),next(NEXT){}
};
Edge first[maxn];
Edge newEdge[maxn*maxn];
int cnt=0;
int m,n;
int level[maxn];
int vis[maxn];
int dist[maxn];
int _LEFT,_RIGHT;
int ans;
int addedge(int from,int to,int cost)
{
	if(first[from].to==0)
		{
			first[from]=Edge(to,cost,0);
		}
	else
		{
			newEdge[++cnt]=first[from];
			first[from]=Edge(to,cost,&newEdge[cnt]);
		}
	return 0;
}
int init()
{
	cnt=0;
	_LEFT=_RIGHT=0;
	ans=INF;
	memset(first,0,sizeof(first));
	memset(newEdge,0,sizeof(newEdge));
	memset(level,0,sizeof(level));
	memset(vis,0,sizeof(vis));
	scanf("%d%d",&m,&n);
	int up;
	int cost,alter,to;
	for(int i=1;i<=n;i++)
		{
			scanf("%d%d%d",&cost,&level[i],&alter);
			addedge(i,n+1,cost);
			for(int j=1;j<=alter;j++)
				{
					scanf("%d%d",&to,&cost);
					addedge(i,to,cost);
				}
		}
	level[n+1]=level[1];
	return 0;
}
int range(int x)
{
	return x>=_LEFT&&x<=_RIGHT;
}
int SPFA(int v0,int vt)
{
	queue <int> Q;
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n+1;i++)
		{
			dist[i]=INF;
		}
	dist[v0]=0;
	vis[v0]=1;
	Q.push(v0);
	while(!Q.empty())
		{
			int now=Q.front();
			Q.pop();
			vis[now]=0;
			for(Edge *p=&first[now];p!=0;p=p->next)
				{
					int to=p->to;
					int cost=p->cost;
					if(range(level[to]))
					if(dist[to]>dist[now]+cost)
						{
							dist[to]=dist[now]+cost;
							if(vis[to]==0)
								{
									vis[to]=1;
									Q.push(to);
								}
						}
				}
		}
	return dist[vt];
	return 0;
}
int solve()
{
	for(_LEFT=level[1]-m,_RIGHT=level[1];_LEFT<=level[1];_LEFT++,_RIGHT++)
		ans=min(ans,SPFA(1,n+1));
	printf("%d\n",ans);
	return 0;
}
int main()
{
	//freopen("1062.in","r",stdin);
	//freopen("1062.out","w",stdout);
	init();
	solve();
	return 0;
}
/*
Tips:这题有陷阱，可能存在比首长高的等级，所以要枚举最短路上的等级范围从[level[1]-m,level[1]]到[level[1],level[1]+m] ！！ 
*/