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: Accepted)

#include<bits/stdc++.h>
#define rep(i,N) for(ll (i)=0;(i)<(N);(i)++)
#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int mod = 1000000007;
const int INF = 1001001001;

int MAX_V = 100; // 頂点数
vector<vector<int>> d(MAX_V, vector<int>(MAX_V, INF));
void warshall_floyd() {
  for(int k = 0; k < MAX_V; ++k)
    for(int i = 0; i < MAX_V; ++i)
      for(int j = 0; j < MAX_V; ++j)  d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> prev(MAX_V, vector<int>(MAX_V, INF));
  vector<int> a(m), b(m), c(m);
  rep(i, m) {
    cin >> a[i] >> b[i] >> c[i];
    --a[i]; --b[i];
    d[a[i]][b[i]] = c[i];
    d[b[i]][a[i]] = c[i];
  }

  rep(i, n) d[i][i] = 0;

  warshall_floyd();

  int ans = m;
  rep(i, m) {
    bool used = false;
    rep(j, n) {
      if(d[j][a[i]] + c[i] == d[j][b[i]]) used = true;
    }
    if(used) ans--;
  }
  cout << ans << endl;
}