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 <cmath>
#include <algorithm>
#include <cstdio>
#include <string>
#include <map>
#include <set>
#include <locale.h>
#include <vector>
#include <iomanip>
#include <cassert>
#include <list>
#include <unordered_map>
#include <cstring>

#define all(x) x.begin(), x.end()
#define sz(x) (int)(x.size())
#define pb push_back

#define onlycin ios_base::sync_with_stdio(false); cin.tie(0)
#define mp make_pair
#define endl "\n"
#define rep(i,x,y) for(int i = x; i <= y; ++i)
#define repr(i,x,y) for(long long i = x; i >= y; --i)
#define y1 wearsdfasdf
#define fi first
#define se second

typedef long long ll;

const int N = (100000) + 322;
const int INF = (ll)(1e9) + 7;

using namespace std;

ll pr[N], a[N], n, pos[N], m, d[111][111], cost[111][111];

struct edge {
  int x, y, cost;
};
vector <edge> e;

bool get(int x, int y, int cost) {
  rep(i, 1, n) {
    rep(j, i, n) {
      if (d[i][j] == d[i][x] + d[y][j] + cost || d[i][j] == d[i][y] + d[j][x] + cost) return 1;
    }
  }
  return 0;
}

int main () {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  rep(i, 1, n) rep(j, 1, n) if (i != j) d[i][j] = INF;
  
  rep(i, 1, m) {
    int x, y, c; cin >> x >> y >> c;
    e.pb({x, y, c});
    d[x][y] = c;
    d[y][x] = c;
  }
  for (int k=1; k<=n; ++k)
    for (int i=1; i<=n; ++i)
      for (int j=1; j<=n; ++j)
        d[i][j] = min (d[i][j], d[i][k] + d[k][j]);
  int cnt = 0;
  for(auto it : e) {
    cnt += get(it.x, it.y, it.cost);
  }
  
  cout << m - cnt;
  return 0;
}