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::cmp::min;

#[allow(unused_macros)]
macro_rules! scan {
    () => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().to_string()
        }
    };
    (;;) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().split_whitespace().map(|s| s.to_string()).collect::<Vec<String>>()
        }
    };
    (;;$n:expr) => {
        {
            (0..$n).map(|_| scan!()).collect::<Vec<_>>()
        }
    };
    ($t:ty) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };
    ($t:ty;;) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.split_whitespace()
                .map(|t| t.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
    ($t:ty;;$n:expr) => {
        (0..$n).map(|_| scan!($t;;)).collect::<Vec<_>>()
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_|
                    scan!($t)
        ).collect::<Vec<_>>()
    };
    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
                    scan!($($t),*)
        ).collect::<Vec<_>>()
    };
}

fn main() {
    let (n, m) = scan!(usize, usize);

    let mut graph = vec![vec![1 << 30; n + 1]; n + 1];
    for i in 0..graph.len() {
        graph[i][i] = 0
    }

    for _ in 0..m {
        let (a, b, c) = scan!(usize, usize, usize);
        graph[a][b] = c;
        graph[b][a] = c;
    }

    let graph0 = graph.clone();
    for k in 1..n + 1 {
        for i in 1..n + 1 {
            for j in 1..n + 1 {
                if graph[i][j] == 1 << 30 || graph[i][k] == 1 << 30 || graph[k][j] == 1 << 30 {
                    continue;
                }
                graph[i][j] = min(graph[i][j], graph[i][k] + graph[k][j])
            }
        }
    }

    let mut ans = 0;
    for i in 1..n + 1 {
        for j in 1..n + 1 {
            if graph0[i][j] != graph[i][j] {
                ans += 1
            }
        }
    }
    println!("{}", ans / 2)
}