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)

#![allow(non_snake_case,unused_imports,dead_code)]
use std::io::Read;
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::BinaryHeap;
use std::cmp::Ordering;

use proconio::input;
use proconio::marker::{Chars,Usize1,Isize1};

fn comp_f64(a: &f64, b: &f64) -> Ordering {
    if a < b {
        return Ordering::Less;
    } else if a > b {
        return Ordering::Greater;
    }
    Ordering::Equal
}

fn main() {
    input!{
        N:usize,
        M:usize,
        ABC:[(Usize1,Usize1,isize);M]
    }

    let INF = 1000000;

    let mut D = WarshallFloyd_initialize(N,INF);
    for &(a,b,c) in &ABC{
        D[a][b] = c;
        D[b][a] = c;
    }
    WarshallFloyd_run(N,&mut D);

    let ans = ABC.iter().filter(|&&(a,b,c)| D[a][b] != c).count();
    println!("{}",ans);
}

fn WarshallFloyd_initialize(N:usize,INF:isize)->Vec<Vec<isize>>{
    let mut Dist = vec![vec![INF;N];N];
    for i in 0..N{
        Dist[i][i] = 0;
    }
    Dist
}
fn WarshallFloyd_run(N:usize,Dist:&mut Vec<Vec<isize>>){
    for k in 0..N{
        for i in 0..N{
            for j in 0..N{
                Dist[i][j] = min(Dist[i][j],Dist[i][k]+Dist[k][j]);
            }
        }
    }
}