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: Wrong Answer)

use proconio::{fastout, input};
use std::io::*;
use std::str::FromStr;

use std::cmp::{max, min};

fn main() {
    input! {
        n: usize,
        m: usize,
        abc: [(usize, usize, i64); m],
    }
    let mut map = vec![vec![std::i64::MAX / 2; n + 1]; n + 1];
    for i in 1..n + 1 {
        map[i][i] = 0;
    }

    for i in 0..abc.len() {
        let x = abc[i];
        map[x.0][x.1] = x.2;
        map[x.1][x.0] = x.2;
    }

    for k in 1..n + 1 {
        for i in 1..n + 1 {
            for j in 1..n + 1 {
                map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
            }
        }
    }

    let mut v = vec![false; m];
    for i in 1..n + 1 {
        for j in 1..n + 1 {
            for k in 0..m {
                let mut x = abc[k];
                if map[i][j] == map[i][x.0] + map[x.1][j] + 1 {
                    v[k] = true;
                }
            }
        }
    }
    println!("{}", v.iter().filter(|&x| *x == false).count());
}