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: Wrong Answer)

use std::io;
use std::collections::BinaryHeap;


fn main() {
    let (n, m) = {
        let i = read::<usize>();
        (i[0], i[1])
    };

    let mut vec = vec![Vec::new(); n];
    let mut cost = vec![vec![0; n]; n];

    for _ in 0..m {
        let (a, b, c) = {
            let i = read::<usize>();
            (i[0] - 1, i[1] - 1, i[2])
        };
        vec[a].push((b, c));
        vec[b].push((a, c));
        cost[a][b] = c;
        cost[b][a] = c;
    }

    let mut dist = vec![Vec::new(); n];
    for j in 0..n {
        dist[j] = dijkstra(&vec, j);
    }

    let mut unused = vec![vec![true; n]; n];
    for i in 0..n { unused[i][i] = false; }
    for i in 0..n { for j in 0..n { for k in 0..n {
        if dist[i][j] + cost[j][k] == dist[i][k] {
            unused[j][k] = false;
        }
    }}}

    let ans = unused.iter().map(|ref v|
        v.iter().filter(|&u| *u == true ).count())
        .sum::<usize>();
    println!("{}", ans / 2);
}


#[allow(dead_code)]
fn read<T>() -> Vec<T>
where T:
    std::str::FromStr,
    T::Err: std::fmt::Debug {

    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    buf.split_whitespace()
        .map(|s| s.trim().parse().unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_one<T>() -> T
where T:
    std::str::FromStr,
    T::Err: std::fmt::Debug {

    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    buf.trim().parse().unwrap()
}

// (pos, cost)
fn dijkstra(adj: &Vec<Vec<(usize, usize)>>, s: usize) -> Vec<usize> {
    let mut dist = (0..adj.len()).map(|_| ::std::usize::MAX)
        .collect::<Vec<usize>>();

    let mut q = BinaryHeap::new();
    dist[s] = 0;
    q.push((s, 0));

    while let Some((pos, cost)) = q.pop() {
        if cost > dist[pos] { continue; }
        for e in &adj[pos] {
            let s = (e.0, e.1 + cost);
            if s.1 < dist[s.0] {
                q.push(s);
                dist[s.0] = s.1;
            }
        }
    }

    dist
}