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)

use std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident) => {
        $it.next().unwrap().parse().unwrap()
    };
    ($it: ident,2) => {
        (parse!($it), parse!($it))
    };
    ($it: ident,3) => {
        (parse!($it), parse!($it), parse!($it))
    };
    ($it: ident,4) => {
        (parse!($it), parse!($it), parse!($it), parse!($it))
    };
    ($it: ident, $n: expr) => {
        (&mut $it).take($n).map(|e| e.parse().unwrap()).collect();
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    let (n, m): (usize, usize) = parse!(it, 2);
    let mut e = vec![];
    let mut d = vec![vec![std::usize::MAX / 10; n]; n];
    for _ in 0..m {
        let (a, b, c): (usize, usize, usize) = parse!(it, 3);
        e.push((a - 1, b - 1, c));
        d[a - 1][b - 1] = c;
        d[b - 1][a - 1] = c;
    }
    for k in 0..n {
        for i in 0..n {
            for j in 0..n {
                d[i][j] = std::cmp::min(d[i][k] + d[k][j], d[i][j]);
            }
        }
    }
    let mut ret = 0;
    for (a, b, c) in e {
        if d[a][b] < c {
            ret += 1;
        }
    }
    println!("{}", ret);
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}