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)

pub mod io {
    use std::io;

    pub fn read_line() -> String {
        let mut s = String::new();
        io::stdin().read_line(&mut s).unwrap();
        s.trim_right().to_owned()
    }

    pub trait Read {
        fn read(s: &str) -> Self;
    }

    macro_rules! def_read {
        ($t: ty) => {
            impl Read for $t {
                fn read(s: &str) -> $t {
                    s.parse().unwrap()
                }
            }
        }
    }

    def_read!(i32);
    def_read!(i64);
    def_read!(isize);
    def_read!(usize);
    def_read!(f64);

    impl<T: Read> Read for Vec<T> {
        fn read(s: &str) -> Self { s.split_whitespace().map(|w| T::read(w)).collect() }
    }

    macro_rules! def_read_tuple {
        ($($t: ident),*) => {
            impl<$($t : Read + Default),*> Read for ($($t),*) {
                fn read(s: &str) -> Self {
                    let ws = s.split_whitespace().collect::<Vec<_>>();
                    let mut i = 0;
                    ( $({i += 1; $t::read(ws[i - 1])}),* )
                }
            }
        };
    }

    def_read_tuple!(T0, T1);
    def_read_tuple!(T0, T1, T2);
    def_read_tuple!(T0, T1, T2, T3);
    def_read_tuple!(T0, T1, T2, T3, T4);

    pub fn readln<T: Read>() -> T { T::read(&read_line()) }
}

fn main() {
    let (n, m) = io::readln();

    let inf = 0x10000000;

    let mut g = vec![vec![inf; n]; n];

    for _ in 0..m {
        let (a, b, c): (usize, usize, i32) = io::readln();
        g[a-1][b-1] = c;
        g[b-1][a-1] = c;
    }

    let og = g.clone();

    for k in 0..n {
        for i in 0..n {
            for j in 0..n {
                g[i][j] = std::cmp::min(g[i][j], g[i][k] + g[k][j]);
            }
        }
    }

    let mut ans = 0;

    for i in 0..n {
        for j in i+1..n {
            if og[i][j] < inf && og[i][j] > g[i][j] {
                ans+=1;
            }
        }
    }

    println!("{}", ans);
}