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)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}



use std::collections::HashSet;
use std::collections::BinaryHeap;

#[derive(PartialEq, Eq, Hash, Clone, Copy)]
struct Edge {
    from: usize,
    to: usize,
    cost: usize,
}

impl PartialOrd for Edge {
    fn partial_cmp(&self, rhs: &Edge) -> Option<std::cmp::Ordering> {
        Some(self.cmp(rhs))
    }
}

impl Ord for Edge {
    fn cmp(&self, rhs: &Edge) -> std::cmp::Ordering {
        self.cost.cmp(&rhs.cost)
    }
}

#[derive(Clone, PartialEq, Eq)]
struct Dist {
    dist: usize,
    edge_from: Option<Edge>
}


impl Dist {
    fn new() -> Dist{
        Dist {
            dist: std::usize::MAX,
            edge_from: None
        }
    }

}

impl PartialOrd for Dist {
    fn partial_cmp(&self, rhs: &Dist) -> Option<std::cmp::Ordering> {
        Some(self.cmp(rhs))
    }
}

impl Ord for Dist {
    fn cmp(&self, rhs: &Dist) -> std::cmp::Ordering {
        rhs.dist.cmp(&self.dist)
    }
}



fn dijkstra(start: usize, edges: &Vec<Vec<Edge>>, set: &mut HashSet<Edge>) {
    let n = edges.len();
    let mut pq = BinaryHeap::new();
    let mut dists = vec![Dist::new(); n];

    dists[start].dist = 0;
    pq.push((0, start));
    while let Some(d) = pq.pop() {
        let v = d.1;
        if dists[v].dist < d.0 {
            continue;
        }
        for e in edges[v].iter() {
            let to = e.to;
            let cost = e.cost;
            let dist = dists[v].dist + cost;
            if dists[to].dist > dist {
                dists[to].dist = dist;
                dists[to].edge_from = Some(*e);
                pq.push((dist, to));
            }
        }
    }

    for d in dists {
        if let Some(e) = d.edge_from {
            let from = e.from;
            let to = e.to;
            let cost = e.cost;
            set.insert(Edge{from: from, to: to, cost: cost});
            set.insert(Edge{from: to, to: from, cost: cost});
        }
    }
}

fn solve(n: usize, es: Vec<(usize, usize, usize)>) -> usize {
    let mut edges = vec![Vec::new(); n];
    for e in es.iter() {
        let from = e.0;
        let to = e.1;
        let cost = e.2;
        edges[from].push(Edge{from: from, to: to, cost: cost});
        edges[to].push(Edge{from: to, to: from, cost: cost});
    }

    let mut used_edges = HashSet::new();
    for i in 0..n {
        dijkstra(i, &edges, &mut used_edges);
    }
    es.len() - used_edges.len() / 2
}




fn main() {
    input! {
        n: usize,
        m: usize,
        es: [(usize1, usize1, usize); m],
    }
    println!("{}", solve(n, es));
}