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: Accepted)

use std::cmp::min;

fn main() {
    let vec = read_line();
    let n = vec[0];
    let m = vec[1];

    let vec = read_lines(m);
    let mut a = vec![];
    let mut b = vec![];
    let mut c = vec![];
    let mut dist = vec![vec![0; 100]; 100];    
    for v in vec {
        a.push(v[0] - 1);
        b.push(v[1] - 1);
        c.push(v[2]);
    }

    // すべての頂点について辞書をつくる
    for i in 0..n as usize {
        for j in 0..n as usize {
            if i == j {
                dist[i][j] = 0;
            } else {
                dist[i][j] = 100000000; // INF
            }
        }
    }

    // 辞書に移動距離を埋める
    for i in 0..m as usize {
        let idx_ai = a[i] as usize;
        let idx_bi = b[i] as usize;
        dist[idx_ai][idx_bi] = min(dist[idx_ai][idx_bi], c[i]);
        dist[idx_bi][idx_ai] = min(dist[idx_bi][idx_ai], c[i]);
    }

    for k in 0..n as usize {
        for i in 0..n as usize {
            for j in 0..n as usize {
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
            }
        }
    }

    // let mut ans = m;
    // for i in 0..m as usize {
    //     let mut shortest = false;
    //     for j in 0..n as usize {
    //         if dist[j][a[i] as usize] + c[i] == dist[j][b[i] as usize] {
    //             shortest = true;
    //         }
    //     }
    //     if shortest {
    //         ans -= 1;
    //     }
    // }
    let mut ans = 0;
    for i in 0..m as usize {
        if dist[a[i] as usize][b[i] as usize] < c[i] {
            ans += 1;
        }
    }

    println!("{}", ans);
}

fn read_line() -> Vec<i32> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    let v = s.trim()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect();
    return v
}

fn read_lines(n : i32) -> Vec<Vec<i32>> {
    let mut v2 = Vec::new();
    for _ in 0..n {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).ok();
        let v = s.trim()
            .split_whitespace()
            .map(|e| e.parse().ok().unwrap())
            .collect();
        v2.push(v);
    }
    v2
}