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::str::FromStr;
use std::cmp::min;
use std::usize;
use std::collections::HashSet;

const INF: i32 = 1000000007;

struct Graph {
    node: Vec<Vec<i32>>,
}

impl Graph {
    fn new(node: usize) -> Graph {
        Graph { node: vec![vec![INF; node]; node] }
    }

    fn add_edge(&mut self, start_node: usize, end_node: usize, edge_cost: i32) -> () {
        self.node[start_node][end_node] = edge_cost;
        self.node[end_node][start_node] = edge_cost;
    }
    fn floyd_warshall(&mut self) -> Vec<Vec<i32>> {
        let mut node = self.node.clone();
        let i = self.node[0].len();
        for relay_node in 0..i {
            for start_node in 0..i {
                for end_node in 0..i {
                    self.node[start_node][end_node] = min(self.node[start_node][end_node], self.node[start_node][relay_node] + self.node[relay_node][end_node]);
                }
            }
        }
        return node;
    }
}

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")
}

fn readc() -> char {
    let s: String = read();
    return s.chars().nth(0).unwrap();
}

fn read_vec<T: FromStr>(n: usize) -> Vec<T> {
    return (0..n).map(|_| read()).collect();
}

fn main() {
    let n: usize = read();
    let m: usize = read();

    let mut g = Graph::new(n+1);
    for i in 0..m {
        let a: usize = read();
        let b: usize = read();
        let c: i32 = read();
        g.add_edge(a, b, c);
    }

    let node = g.floyd_warshall();
    let mut ans = 0;
    for i in 1..n+1 {
        for j in 1..n+1 {
            if i == j {
                continue;
            }
            // println!("{} {} {} {}", i, j, node[i][j], g.node[i][j]);
            if node[i][j] != INF && node[i][j] != g.node[i][j] {
                ans += 1;
            }
        }
    }
    println!("{}", ans / 2);
}