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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
#![allow(unused_mut)]
#![allow(dead_code)]
#![allow(unused_variables)]

use std::collections::HashSet;
use std::collections::HashMap;
use std::collections::BTreeSet;
use std::collections::VecDeque;
use std::cmp::{max, min};
use std::mem::swap;
use std::io::prelude::*; // input()

fn input<T>() -> T
    where T: std::str::FromStr {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}
static INF: usize = 1_000_000;

fn main() {
    let n = input::<usize>();
    let m = input::<usize>();
    let abc = (0..m)
        .map(|_| (input::<usize>()-1,input::<usize>()-1,input::<usize>()))
        .collect::<Vec<_>>();

    let mut d = vec![vec![INF; 100]; 100];
    for i in 0..n {
        d[i][i] = 0;
    }
    for &(a, b, c) in &abc {
        d[a][b] = min(d[a][b], c);
        d[b][a] = min(d[b][a], c);
    }

    for k in 0..n {
        for i in 0..n {
            for j in 0..n {
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
            }
        }
    }

    let mut ans = m;

    for &(a, b, c) in &abc {
        let shortest = || {
            for j in 0..n {
                if d[j][a] + c == d[j][b]  {
                   return true;
                }
            }
            false
        };
        if shortest() {
            ans-=1;
        }
    }
    println!("{}", ans);
}