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::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

use std::collections::VecDeque;

fn main() {
    let n: usize = read();
    let m: usize = read();
    // <to, weight, id>
    let mut map: Vec<Vec<(usize, usize, usize)>> = vec![vec![]; n + 1];
    let mut mv: Vec<(usize, usize, usize)> = vec![];

    for i in 0..m {
        let a: usize = read();
        let b: usize = read();
        let c: usize = read();
        map[a].push((b, c, i));
        map[b].push((a, c, i));
        mv.push((a, b, c));
    }

    let mut mapd: Vec<Vec<usize>> = vec![vec![0; n + 1]; n + 1];
    for i in 1..n + 1 {
        // cn, total_weight, edge_id
        let mut vd: VecDeque<(usize, usize, usize)> = VecDeque::new();
        vd.push_front((i, 0, m));
        let mut memo = vec![1_000_000_000; n + 1];
        let mut is_done = vec![false; m];
        loop {
            match vd.pop_front() {
                None => break,
                Some(s) => {
                    // println!("{:?}", s);
                    if s.2 < m {
                        is_done[s.2] = true;
                    }
                    if s.1 < memo[s.0] {
                        memo[s.0] = s.1;
                    }
                    for i in &map[s.0] {
                        if !is_done[i.2] {
                            vd.push_back((i.0, memo[s.0] + i.1, i.2))
                        }
                    }
                }
            }
        }
        for j in 1..n + 1 {
            mapd[i][j] = memo[j];
        }
    }

    let mut cnt = 0;
    for mi in mv {
        let mut flag = false;
        'outer: for i in 1..n + 1 {
            for j in 1..n + 1 {
                if mapd[i][mi.0] + mi.2 + mapd[mi.1][j] == mapd[i][j] {
                    flag = true;
                    break 'outer;
                }
            }
        }
        if !flag {
            cnt += 1;
        }
    }
    println!("{}", cnt);
}