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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
use proconio::marker::Usize1;
use std::collections::{VecDeque, BinaryHeap};
use std::cmp::Reverse;

fn main()
{
    input! {
    n:usize,
    m:usize,
    edge:[(Usize1,Usize1,i64);m],
    }

    let mut node = vec![vec![]; n];
    for (i, &(a, b, c)) in edge.iter().enumerate() {
        node[a].push((b, c, i));
        node[b].push((a, c, i));
    }

    let inf = 1i64 << 60;
    let mut used = vec![false; m];
    for i in 0..n {
        let mut dist = vec![inf; n];
        dist[i] = 0;
        let mut q = BinaryHeap::new();
        q.push(Reverse((0, i)));
        while let Some(Reverse((cost, idx))) = q.pop() {
            if dist[idx] < cost {
                continue;
            }
            for &(nxt, c, _) in node[idx].iter() {
                let cst = c + cost;
                if dist[nxt] <= cst {
                    continue;
                }
                dist[nxt] = cst;
                q.push(Reverse((cst, nxt)));
            }
        }

        for j in (i + 1)..n {
            let mut q = VecDeque::new();
            q.push_back(j);
            while let Some(idx) = q.pop_front() {
                for &(nxt, cst, index) in node[idx].iter() {
                    if dist[idx] != dist[nxt] + cst {
                        continue;
                    }
                    q.push_back(nxt);
                    used[index] = true;
                }
            }
        }
    }

    let ans = used.into_iter().filter(|&n| !n).count();
    println!("{}", ans);
}