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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MOD = 1000000007;
const int INF = 1e15;
//using Graph = vector<vector<int>>;

signed main(){
  int N, M;
  cin >> N >> M;

  vector<vector<int>> G(N, vector<int>(N, INF));
  vector<pair<int, int>> E(M);
  vector<int> cost(M);
  for( int i = 0; i < N; i++ ) G.at(i).at(i) = 0;
  for( int i = 0; i < M; i++ ){
    int A, B, C;
    cin >> A >> B >> C;
    A--; B--;
    G.at(A).at(B) = C;
    G.at(B).at(A) = C;
    E.at(i) = {A, B};
    cost.at(i) = C;
  }

  for (int k = 0; k < N; k++){       // 経由する頂点
    for (int i = 0; i < N; i++) {    // 始点
      for (int j = 0; j < N; j++) {  // 終点
        G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
      }
    }
  }

  vector<bool> fg(M, true);
  for( int i = 0; i < M; i++ ){
    for( int s = 0; s < N; s++ ){
      for( int t = s+1; t < N; t++ ){
        //s, t = 最短経路を見る二頂点
        int A = G[s][E.at(i).first] + cost.at(i) + G[E.at(i).second][t];
        if( A == G[s][t] ) fg.at(i) = false;
      }
    }
  }
  int ans = 0;
  for( int i = 0; i < M; i++ ){
    if( fg.at(i) ) ans++;
  }
  cout << ans << endl;

 }