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::collections::HashSet;

struct Graph {
    edges: Vec<Vec<(usize, i64)>>, // adjacent list
    n: usize
}

impl Graph {
    fn new(n: usize) -> Self {
        Graph {
            edges: vec![Vec::new(); n],
            n: n
        }
    }

    fn add_costs(&mut self, from: usize, to: usize, cost: i64, directed: bool) {
        if directed {
            self.edges[from].push((to, cost));
        } else {
            self.edges[from].push((to, cost));
            self.edges[to].push((from, cost));
        }

    }

    fn dijkstra_search(&self, start: usize) -> (Vec<i64>, Vec<usize>) {
        const INF: i64 = 1 << 61;
        let mut dist: Vec<i64> = vec![INF; self.n]; // Dijkstraなのでi64ではなくusizeの方がいいかも
        let mut visited: Vec<bool> = vec![false; self.n];
        let mut prev: Vec<usize> = vec![INF as usize; self.n]; //nodeを格納するのにusizeでないの気持ち悪い
        dist[start] = 0;
        let mut que: BinaryHeap<(i64, usize)> = BinaryHeap::new();
        que.push((0, start));
        while let Some((_, v)) = que.pop() {
            if visited[v] {
                continue
            }
            visited[v] = true;
            for edge in &self.edges[v] {
                let (u, cost) = *edge;
                if visited[u] {
                    continue
                }
                if dist[u] > dist[v] + cost {
                    dist[u] = dist[v] + cost;
                    prev[u] = v;
                    que.push(( -dist[u], u));
                }
            }
        }
        return (dist, prev)
    }

    fn dijkstra_shortest_path(&self, start: usize, goal: usize) -> Vec<usize> {
        const INF: usize = 1 << 61;
        let (dist, prev) = self.dijkstra_search(start);
        let mut shortest_path: Vec<usize>= vec![];
        let mut node = goal;
        while node != INF {
            shortest_path.push(node);
            node = prev[node];
        }
        shortest_path.reverse();
        return shortest_path
    }
}

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input! {
    n: usize, m: usize,
    abc: [(usize, usize, i64); m],
    }
    let mut graph = Graph::new(n);
    for i in 0..m {
        graph.add_costs(abc[i].0 - 1, abc[i].1 - 1, abc[i].2, false);
    }
    let mut used_paths: Vec<HashSet<usize>> = vec![HashSet::new(); n];
    for i in 0..n {
        for j in i+1..n {
            let path = graph.dijkstra_shortest_path(i, j);
            if path.len() == 2 {
                used_paths[path[0]].insert(path[1]);
                used_paths[path[1]].insert(path[0]);
            } else {
                let l = used_paths.len();
                for k in 0..l-1 {
                    used_paths[path[k]].insert(path[k+1]);
                    used_paths[path[k+1]].insert(path[k]);
                }
            }
        }
    }
    let l = used_paths.len();
    let mut ans = 0;
    for i in 0..l {
        ans += used_paths[i].len();
    }
    println!("{}", m - ans/2);
}