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: Accepted)

#include <stdio.h>
#define INF 1 << 29

int min(int a,int b){
  if(a < b){
    return a;
  }
  return b;
}

int main(void){
  int N,M;
  scanf("%d%d",&N,&M);

  int a[M],b[M],c[M],dist[N][N];

  for(int i = 0;i < N;i++){
    for(int j = 0;j < N;j++){
      if(i == j)
        dist[i][j] = 0;
      else
        dist[i][j] = INF;
    }
  }

  for(int i = 0;i < M;i++){
    scanf("%d%d%d",&a[i],&b[i],&c[i]);
    a[i]--; b[i]--;
    dist[a[i]][b[i]] = dist[b[i]][a[i]] = c[i];
  }

  for(int k = 0;k < N;k++){
    for(int i = 0;i < N;i++){
      for(int j = 0;j < N;j++){
        dist[i][j] = min(dist[i][j],dist[i][k]+dist[k][j]);
      }
    }
  }

  int ans = 0;
  for(int i = 0;i < M;i++){
    if(dist[a[i]][b[i]] < c[i]) ans++;
  }

  printf("%d\n",ans);

  return 0;
}