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 Rust (1.15.1) (judge verdict: Accepted)

fn main() {
    let (n,m) = {
        let s = rv::<usize>();
        (s[0],s[1])
    };
    let mut g=vec![vec![];n];
    for i in 0..m {
        let s = rv::<usize>();
        let (a,b,c) = (s[0]-1,s[1]-1,s[2]as i32);
        g[a].push((b,c,i));
        g[b].push((a,c,i));
    }
    let mut e=vec![1;m+1];
    let mut q=std::collections::BinaryHeap::new();
    let mut d=vec![INF;n];
    for i in 0..n {
        for v in d.iter_mut() {*v=INF}
        q.push((INF-0, i, m));
        while let Some((s,v,i))=q.pop() {
            let s=-s;
            if d[v]>=s {
                e[i]=0;
            }
            if d[v]<=s {
                continue;
            }
            d[v]=s;
            for &(u,c,i) in &g[v] {
                let s=s+c;
                if d[u]>=s {
                    q.push((-s,u,i));
                }
            }
        }
    }
    println!("{}",e.into_iter().sum::<i32>());
}
const INF: i32=std::i32::MAX;
fn rl() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s
}
fn rv<T>() -> Vec<T>
  where T: std::str::FromStr, T::Err: std::fmt::Debug, {
    rl().split_whitespace().map(|x|x.parse::<T>().unwrap()).collect()
}