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Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his house is on the other. A nice walk through the forest, seeing the birds and chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different route everyday. He also wants to get home before dark, so he always takes a path to make progress towards his house. He considers taking a path from A to B to be progress if there exists a route from B to his home that is shorter than any possible route from A. Calculate how many different routes through the forest Jimmy might take.

Input

Input contains several test cases followed by a line containing 0. Jimmy has numbered each intersection or joining of paths starting with 1. His office is numbered 1, and his house is numbered 2. The first line of each test case gives the number of intersections N, 1 < N ≤ 1000, and the number of paths M. The following M lines each contain a pair of intersections a b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between intersection a and a different intersection b. Jimmy may walk a path any direction he chooses. There is at most one path between any pair of intersections.

Output

For each test case, output a single integer indicating the number of different routes through the forest. You may assume that this number does not exceed 2147483647

Sample Input

5 6
1 3 2
1 4 2
3 4 3
1 5 12
4 2 34
5 2 24
7 8
1 3 1
1 4 1
3 7 1
7 4 1
7 5 1
6 7 1
5 2 1
6 2 1
0
output
2
4








代碼如下：源代碼
  1 #include<stdio.h>
  2 #include<string.h>
  3 #define INF 0x3f3f3f3f
  4 #define M 1010
  5 #define typec int 
  6 int map[M][M];
  7 int mark[M];
  8 int dist[M];
  9 int sum[M];
 10 int n,m;
 11 int path[M];
 12 int dijkstra()
 13 {
 14     int i,j;
 15     int k;
 16     int min;
 17     for(i=1;i<=n;i++)
 18     {
 19         dist[i]=map[2][i];
 20         mark[i]=0;
 21         path[i]=2;
 22     }
 23     dist[2]=0;//從2開始
 24     mark[2]=1;
 25     path[2]=-1;
 26 
 27 
 28     for(j=2;j<n;j++)//注意不要寫錯成i++
 29     {
 30     min=INF;
 31     for(i=1;i<=n;i++)
 32     {
 33         if(!mark[i]&&min>dist[i])
 34         {
 35             min=dist[i];
 36             k=i;
 37         }
 38     
 39     }
 40     if(min>=INF)//注意要在循環(huán)外面
 41             break;
 42     
 43     mark[k]=1;
 44     for(i=1;i<=n;i++)
 45     {
 46         if(!mark[i]&&dist[i]>dist[k]+map[k][i])
 47             dist[i]=dist[k]+map[k][i],path[i]=k;
 48 
 49     }
 50     }
 51     //printf("%d\n",min);
 52 return min;
 53 
 54 }
 55 int dfs(int i,int n)//記憶性搜索，類似于動態(tài)規(guī)劃的方法，記錄下來
 56  {    if(i==2)  return 1;  
 57   if(sum[i]!=-1) 
 58       return sum[i];    
 59   int cnt=0;    
 60   for(int j=1;j<=n;j++)  
 61   {      
 62       if(map[i][j]<INF&&dist[j]<dist[i])  
 63           cnt+=dfs(j,n);
 64     }    sum[i]=cnt;  
 65   return sum[i]; 
 66   }    
 67 int main()
 68 {    int i,j;
 69     //init();
 70     while(scanf("%d%d",&n,&m)&&n!=0)
 71     {
 72         int a,b,d;
 73         for(i=1;i<=n;i++)    
 74        for(j=1;j<=n;j++) //更新數(shù)據(jù)      
 75        {   
 76            if(i==j)
 77                map[i][j]=0; 
 78            else map[i][j]=INF;  
 79        } 
 80         int s;
 81         for(i=0;i<m;i++)
 82         {
 83             scanf("%d %d %d",&a,&b,&d);
 84             if(map[a][b]>=d)
 85             map[a][b]=map[b][a]=d;
 86         }
 87         
 88     //     Dijkstra(map,dist,n,2);
 89          dijkstra();
 90          memset(sum,-1,sizeof(sum));//更新數(shù)據(jù)
 91         s=dfs(1,n);
 92         printf("%d\n",s);
 93     }
 94     
 95     return 0;
 96 }
 97 
 98 
 99 
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