Problem:
Score : 400 points

Problem Statement

You are given an undirected connected weighted graph with N vertices and M edges that contains neither self-loops nor double edges.

The i-th (1≤i≤M) edge connects vertex a_i and vertex b_i with a distance of c_i.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

A connected graph is a graph where there is a path between every pair of different vertices.

Find the number of the edges that are not contained in any shortest path between any pair of different vertices.

Constraints

2≤N≤100

N-1≤M≤min(N(N-1)/2,1000)

1≤a_i,b_i≤N

1≤c_i≤1000

c_i is an integer.

The given graph contains neither self-loops nor double edges.

The given graph is connected.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M

Output

Print the number of the edges in the graph that are not contained in any shortest path between any pair of different vertices.

Sample Input 1

3 3
1 2 1
1 3 1
2 3 3

Sample Output 1

1

In the given graph, the shortest paths between all pairs of different vertices are as follows:

The shortest path from vertex 1 to vertex 2 is: vertex 1 → vertex 2, with the length of 1.

The shortest path from vertex 1 to vertex 3 is: vertex 1 → vertex 3, with the length of 1.

The shortest path from vertex 2 to vertex 1 is: vertex 2 → vertex 1, with the length of 1.

The shortest path from vertex 2 to vertex 3 is: vertex 2 → vertex 1 → vertex 3, with the length of 2.

The shortest path from vertex 3 to vertex 1 is: vertex 3 → vertex 1, with the length of 1.

The shortest path from vertex 3 to vertex 2 is: vertex 3 → vertex 1 → vertex 2, with the length of 2.

Thus, the only edge that is not contained in any shortest path, is the edge of length 3 connecting vertex 2 and vertex 3, hence the output should be 1.

Sample Input 2

3 2
1 2 1
2 3 1

Sample Output 2

0

Every edge is contained in some shortest path between some pair of different vertices.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(){
  int n,m,a,b,c,i,j,k,s=0,d[110][110],f[110][110],e[110][110]={};
  int u[110]={};
  scanf("%d %d",&n,&m);
  for(i=0;i<n*n;i++)d[i/n][i%n]=1e9;
  for(i=0;i<n*n;i++)e[i/n][i%n]=-1;
  for(i=0;i<n*n;i++)f[i/n][i%n]=-1;
  for(i=0;i<n;i++)d[i][i]=0;
  for(i=0;i<n;i++)f[i][i]=i;
  for(i=0;i<m;i++){
    scanf("%d %d %d",&a,&b,&c);
    a--;
    b--;
    d[a][b]=d[b][a]=c;
    e[a][b]=e[b][a]=i;
    f[a][b]=a;
    f[b][a]=b;
  }
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      for(k=0;k<n;k++){
	if(d[j][k]>d[j][i]+d[i][k]){
	  d[j][k]=d[j][i]+d[i][k];
	  f[j][k]=f[i][k];
	  e[j][k]=e[i][k];
	}
      }
    }
  }
  /*printf("\n");
  for(i=0;i<n;i++){
    for(j=0;j<n;j++)printf("%d ",f[i][j]);printf("\n");
    }
  printf("\n");
  for(i=0;i<n;i++){
    for(j=0;j<n;j++)printf("%d ",e[i][j]);printf("\n");
    }//*/
 for(i=0;i<n;i++){
    for(j=i+1;j<n;j++){
      s+=1-u[e[i][j]];
      u[e[i][j]]=1;
    }
  }
  printf("%d\n",m-s);
  return 0;
}