G - The Shortest Path in Nya Graph

This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just solo hay que cambiar un poco el algoritmo. If you do not understand a word of this paragraph, just move on.
The Nya graph is an undirected graph with “layers”. Each node in the graph belongs to a layer, there are N nodes in total.
You can move from any node in layer x to any node in layer x + 1, with cost C, since the roads are bi-directional, moving from layer x + 1 to layer x is also allowed with the same cost.
Besides, there are M extra edges, each connecting a pair of node u and v, with cost w.
Help us calculate the shortest path from node 1 to node N.
Input
The first line has a number T (T <= 20) , indicating the number of test cases.
For each test case, first line has three numbers N, M (0 <= N, M <= 10 5) and C(1 <= C <= 10 3), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers l i (1 <= l i <= N), which is the layer of i th node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 10 4), which means there is an extra edge, connecting a pair of node u and v, with cost w.
Output
For test case X, output "Case #X: " first, then output the minimum cost moving from node 1 to node N.
If there are no solutions, output -1.
Sample Input
2
3 3 3
1 3 2
1 2 1
2 3 1
1 3 3

3 3 3
1 3 2
1 2 2
2 3 2
1 3 4
Sample Output
Case #1: 2
Case #2: 3

找了半天错结果是 ++cnt 写成了 cnt++.

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#define N 1000005
#define INF 0x3f3f3f3f
using namespace std;
int cnt;
int lay[N];  //层数 
int head[N]; //结点连接的一条边
int dis[N];  //记录最短距离 
bool inq[N];  
struct edge{
	int v,w,next;  // 目标  权重  下一个 
}edge[N]; 
queue<int> q;
void Addedge(int u,int v,int w){
	edge[++cnt].v=v;
	edge[cnt].w=w;
	edge[cnt].next=head[u];
	head[u]=cnt;
}
void Initial(){
	memset(lay,-1,sizeof(lay));
	memset(head,-1,sizeof(head));
	cnt=0;
	memset(dis,INF,sizeof(dis));
	memset(inq,0,sizeof(inq));
	
}
void SPFA(){
	int x=1;
	q.push(x);
	inq[x]=1;
	dis[x]=0;
	while(q.size())
	{
		x=q.front();
		q.pop();
		inq[x]=0;
		for(int i=head[x];i!=-1;i=edge[i].next){
			int w=edge[i].w,v=edge[i].v;
			if( dis[v]> dis[x]+ w){
				dis[v]= dis[x]+w;
				if(!inq[v]){
					q.push(v);
					inq[v]=1;
				}
			}
		
		}		
	}

}
int main()
{
	int t;
	scanf("%d",&t);
	for(int tt=1;tt<=t;tt++){
		Initial();

		int n,m,c;
		scanf("%d %d %d",&n,&m,&c);
		for(int i=1;i<=n;i++){
			scanf("%d",&lay[i]);
		} 
		for(int i=1;i<=n;i++){
			Addedge(n+lay[i],i,0);
			if(lay[i]>1)
				Addedge(i,n+lay[i]-1,c);
			if(lay[i]<n)
				Addedge(i,n+lay[i]+1,c);		
		}
		for(int i=1;i<=m;i++){
			int u,v,w;
			scanf("%d %d %d",&u,&v,&w);
			Addedge(u,v,w);
			Addedge(v,u,w);
		}
		//建图完成
		 SPFA();
		if(dis[n]==INF) dis[n]=-1;
		printf("Case #%d: %d\n",tt,dis[n]);
	}
	
	return 0;
}