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 G = GraphWarshallFloyd::new(N,INF);

    for &(a,b,c) in &ABC{
        G.set(a,b,c);
        G.set(b,a,c);
    }
    G.run();

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

struct GraphWarshallFloyd{
    N:usize,
    Dist:Vec<Vec<isize>>,
    INF:isize
}

impl GraphWarshallFloyd {
    fn new(N:usize,INF:isize) -> GraphWarshallFloyd{
        let mut ret = GraphWarshallFloyd{N:N,Dist:Vec::new(),INF:INF};
        ret.Dist = vec![vec![INF;N];N];
        for i in 0..N{
            ret.Dist[i][i] = 0;
        }
        ret
    }
    fn set(&mut self,from:usize,to:usize,dist:isize){
        self.Dist[from][to] = dist;
    }
    fn get(&self,from:usize,to:usize)->isize{
        self.Dist[from][to]
    }
    fn run(&mut self){
        let N = self.N;
        for k in 0..N{
            for i in 0..N{
                for j in 0..N{
                    self.Dist[i][j] = min(self.Dist[i][j],self.Dist[i][k]+self.Dist[k][j]);
                }
            }
        }
    }
}