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: Runtime Error)

use std::collections::BinaryHeap;
use std::cmp::Ordering;

#[derive(Clone)]
struct Edge {
    to: usize,
    cost: usize
}

// https://atcoder.jp/contests/abc021/submissions/5777748
// ↓
use std::io::prelude::*;

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()
}
// ↑
// https://atcoder.jp/contests/abc021/submissions/5777748

// https://qiita.com/hatoo@github/items/fa14ad36a1b568d14f3e
// ↓
#[derive(Eq, PartialEq, Clone, Debug)]
pub struct Rev<T>(pub T);

impl<T: PartialOrd> PartialOrd for Rev<T> {
    fn partial_cmp(&self, other: &Rev<T>) -> Option<Ordering> {
        other.0.partial_cmp(&self.0)
    }
}

impl<T: Ord> Ord for Rev<T> {
    fn cmp(&self, other: &Rev<T>) -> Ordering {
        other.0.cmp(&self.0)
    }
}
// ↑
// https://qiita.com/hatoo@github/items/fa14ad36a1b568d14f3e

fn main() {
    let n = input::<usize>(); // 頂点数
    let a = input::<usize>() - 1; // スタート地点
    let b = input::<usize>() - 1; // 目的地
    let m = input::<usize>(); // パス数
    let xys = (0..m)
        .map(|_| (input::<usize>()-1, input::<usize>()-1))
        .collect::<Vec<_>>();

    let mut d = vec![usize::max_value(); n];
    d[a] = 0;

    let mut g = vec![vec![]; n];
    for (x, y) in xys {
        g[x].push(Edge {to: y, cost: 1});
        g[y].push(Edge {to: x, cost: 1});
    }

    let mut que = BinaryHeap::new();
    que.push(Rev((0, a)));

    while !que.is_empty() {
        let Rev((v_min_d, v)) = que.pop().unwrap();
        if d[v] < v_min_d {
            continue;
        }
        for e in &g[v] {
            if d[e.to] > d[v] + e.cost {
                d[e.to] = d[v] + e.cost;
                que.push(Rev((d[e.to], e.to)))
            }
        }
    }

    println!("スタート地点から各地点までの最短距離");
    for i in d.iter().enumerate() {
        println!("{:?} ", i);
    }
    println!("目的地までの最短距離: {}", d[b]);
}