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)

#[macro_use]
mod input {

use std;
use std::io;

const SPLIT_DELIMITER: char = ' ';

#[macro_export]
#[allow(unused_macros)]
macro_rules! input {
    ( $($x:expr ),*) => {
        {
            let temp_str = input_line_str();
            let mut split_result_iter = temp_str.split_whitespace();
                $(
                let buf_split_result = split_result_iter.next();
                let buf_split_result = buf_split_result.unwrap();
                    ($x) = buf_split_result.parse().unwrap();
                )*
        }
    };
}

#[allow(dead_code)]
pub fn input_line_str() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

#[allow(dead_code)]
pub fn p<T>(t: T)
where
    T: std::fmt::Display,
{
    println!("{}", t);
}

#[allow(dead_code)]
pub fn input_vector2d<T>(line: usize) -> Vec<Vec<T>>
where
    T: std::str::FromStr,
{
    let mut v: Vec<Vec<T>> = Vec::new();

    for _ in 0..line {
        let vec_line = input_vector();
        v.push(vec_line);
    }
    v
}

#[allow(dead_code)]
pub fn input_vector<T>() -> Vec<T>
where
    T: std::str::FromStr,
{
    let mut v: Vec<T> = Vec::new();

    let s = input_line_str();
    let split_result = s.split(SPLIT_DELIMITER);
    for z in split_result {
        let buf = match z.parse() {
            Ok(r) => r,
            Err(_) => panic!("Parse Error"),
        };
        v.push(buf);
    }
    v
}

#[allow(dead_code)]
pub fn str2vec(s: &str) -> Vec<char> {
    let mut v: Vec<char> = Vec::new();
    for c in s.chars() {
        v.push(c);
    }
    v
}
}

mod graph {

use std;
use std::cmp;
use std::collections::BinaryHeap;

pub fn bellman_ford(graph: &Vec<Vec<(usize, isize)>>, dist: &mut Vec<Option<isize>>) -> bool {
    //負の閉路検出用
    let mut neg_flag = false;
    for _ in 0..graph.len() {
        neg_flag = false;
        for (from, v) in graph.iter().enumerate() {
            for e in v {
                let cost = e.1;
                let to = e.0;
                if let Some(x) = dist[from] {
                    if let None = dist[to] {
                        dist[to] = Some(x + cost);
                    } else if let Some(y) = dist[to] {
                        if y > x + cost {
                            neg_flag = true;
                            dist[to] = Some(x + cost);
                        }
                    }
                }
            }
        }
    }
    neg_flag
}

pub fn dijkstra(graph: &Vec<Vec<(usize, isize)>>, start: usize) -> Vec<Option<isize>> {
    let mut heap = BinaryHeap::new();
    let mut dist = vec![None; graph.len()];
    dist[start] = Some(0);
    for e in &graph[start] {
        heap.push((dist[start].unwrap() - 1 * e.1, e.0));
    }

    while !heap.is_empty() {
        let e = heap.pop().unwrap();
        if let Some(x) = dist[e.1] {
            if x < -1 * e.0 {
                continue;
            }
        }
        dist[e.1] = Some(-1 * e.0);
        for next_e in &graph[e.1] {
            if dist[next_e.0] == None {
                heap.push(((-1 * (dist[e.1].unwrap() + next_e.1)), next_e.0));
            }
        }
    }
    dist
}

pub fn warshall_floyd(graph: &mut Vec<Vec<isize>>) {
    for k in 0..graph.len() {
        for i in 0..graph.len() {
            for j in 0..graph.len() {
                graph[i][j] = cmp::min(graph[i][j], graph[i][k] + graph[k][j]);
            }
        }
    }
}
}

#[allow(unused_imports)]
use graph::*;
#[allow(unused_imports)]
use input::*;

fn main() {
    let (n, m): (usize, usize);
    input!(n, m);
    let mut v = vec![vec![]; n];
    let mut input_str = vec![];
    for _ in 0..m {
        let (a, b, c): (usize, usize, isize);
        input!(a, b, c);
        input_str.push((a, b, c));
        v[a - 1].push((b - 1, c));
        v[b - 1].push((a - 1, c));
    }

    let mut ans = 0;

    for &(a, b, c) in &input_str {
        let s1 = dijkstra(&v, a - 1);
        let i1 = v[a - 1].iter().position(|&x| x == (b - 1, c)).unwrap();
        let i2 = v[a - 1].iter().position(|&x| x == (b - 1, c)).unwrap();
        v[a - 1].remove(i1);
        v[b - 1].remove(i2);
        let s2 = dijkstra(&v, a - 1);
        if s1 == s2 {
            ans += 1;
        }
        v[a - 1].push((b - 1, c));
        v[b - 1].push((a - 1, c));
    }
    p(ans);
}