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)

use std::io;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::f64::consts::PI;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}


fn main() {
    let line: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
    let n = line[0];
    let m = line[1];
    let mut g = vec![Vec::new(); n];
    let mut e = Vec::new();
    for i in 0..m {
        let line: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
        let a = line[0] - 1;
        let b = line[1] - 1;
        let c = line[2] as i64;

        g[a].push((b, i));
        g[b].push((a, i));
        e.push(c)
    }

    let mut used = vec![false; m+1];
    for i in 0..n {
        let mut que = BinaryHeap::new();
        que.push((0 as i64, i, m));
        let mut memo = vec![n; n];
        let mut cost = vec![1_000_010; n];
        while !que.is_empty() {
            let (cur, x, ei) = que.pop().unwrap();
            if cost[x] < -cur {
                continue;
            }
            used[ei] = true;
            if cost[x] == -cur {
                continue;
            }
            cost[x] = -cur;

            for i in 0..g[x].len() {
                let (nxt, ei) = g[x][i];
                que.push((cur - e[ei], nxt, ei));
            }
        }
    }
    let ans = used.iter().fold(0, |acc, val| if !*val { acc + 1 } else { acc });

    println!("{}", ans);
}