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 <iostream>
#include <algorithm>
using namespace std;
#define MAX 110
#define MAX_E 1010
#define INF 10000000


int e[MAX][MAX];
int d[MAX][MAX];
int n;

void warshall_floyd() {
  for (int k = 0; k < n; k++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (d[i][j] > d[i][k] + d[k][j]) {
          d[i][j] = d[i][k] + d[k][j];
        }
      }
    }
  }
}

bool contains(int i, int j) {
  for (int s = 0; s < n; s++) {
    if (s == j) continue;
    if (d[s][i] + e[i][j] == d[s][j]) return true;
  }
  return false;
}

int main() {
  for (int i = 0; i < MAX; i++) {
    for (int j = 0; j < MAX; j++) {
      if (i == j) d[i][j] = 0;
      else d[i][j] = INF;
    }
  }
  int m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) e[i][j] = 0;
      else e[i][j] = INF;
    }
  }
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    e[a - 1][b - 1] = c;
    e[b - 1][a - 1] = c;
    d[a - 1][b - 1] = c;
    d[b - 1][a - 1] = c;
  }
  warshall_floyd();
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (e[i][j] != INF && e[i][j] > d[i][j]) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}