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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

const INF: i64 = 1000 * 10000;

fn warshall_floyd(n: usize, mut edges: Vec<Vec<i64>>) -> usize {
    let mut prev = vec![vec![0; n]; n];
    for i in 0..n {
        for j in 0..n {
            prev[i][j] = i;
        }
    }
    for k in 0..n {
        for i in 0..n {
            for j in 0..n {
                if edges[i][k] + edges[k][j] < edges[i][j] {
                    edges[i][j] = edges[i][k] + edges[k][j];
                    prev[i][j] = prev[k][j];
                }
            }
        }
    }
    let mut used = HashSet::new();
    for start in 0..n {
        for goal in 0..n {
            let mut cur = goal;
            while cur != start {
                used.insert((min(prev[start][cur], cur), max(prev[start][cur], cur)));
                cur = prev[start][cur];
            }
        }
    }
    used.len()
}

fn main() {
    let v = read_vec::<usize>();
    let (n, m) = (v[0], v[1]);
    let mut edges: Vec<Vec<i64>> = vec![vec![INF; n]; n];
    for i in 0..m {
        let v = read_vec::<usize>();
        let (a, b, c) = (v[0] - 1, v[1] - 1, v[2] as i64);
        edges[a][b] = c;
        edges[b][a] = c;
    }
    println!("{}", m - warshall_floyd(n, edges));
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}