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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::input;
use proconio::marker::Usize1;

fn adjacency_list(n: usize, uvc: &Vec<(usize, usize, u64)>) -> Vec<Vec<(usize, u64)>> {
    let mut e = vec![vec![]; n];
    for &(u, v, c) in uvc.iter() {
        e[u].push((v, c));
        e[v].push((u, c));
    }
    e
}

fn dijkstra(n: usize, inf: u64, e: &Vec<Vec<(usize, u64)>>, s: usize) -> Vec<u64> {
    let mut d = vec![inf; n];
    let mut pq = std::collections::BinaryHeap::new();
    d[s] = 0;
    pq.push(std::cmp::Reverse((0, s)));
    while let Some(std::cmp::Reverse((c_u, u))) = pq.pop() {
        if c_u > d[u] {
            continue;
        }
        for &(v, c_v) in e[u].iter() {
            let c = c_u + c_v;
            if c < d[v] {
                d[v] = c;
                pq.push(std::cmp::Reverse((c, v)));
            }
        }
    }
    d
}

fn main() {
    input! {
        n: usize,
        m: usize,
        abc: [(Usize1, Usize1, u64); m],
    };
    let inf = 1_000_000_000;
    let e = adjacency_list(n, &abc);
    // d[u][v]: u から v への最短距離
    let mut d = vec![];
    for u in 0..n {
        d.push(dijkstra(n, inf, &e, u));
    }
    let mut ans = 0;
    for &(u, v, c) in abc.iter() {
        if d[u][v] != c {
            ans += 1;
        }
    }
    println!("{}", ans);
}
*/

fn main() {
    let exe = "/tmp/bin6399BCAB";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";