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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

int min(int a, int b){
    if(a < b){
        return a;
    }else{
        return b;
    }
}

int main(void){

    int N, M;

    scanf("%d%d", &N, &M);

    int a[M], b[M], c[M];
    int dist[N][N];

    for(int i = 0; i < M; i++){
        scanf("%d%d%d", &a[i], &b[i], &c[i]);
        a[i]--;
        b[i]--;
    }

    for(int i = 0; i < N; i++){
        for(int j = 0; j < N; j++){
            if(i == j){
                dist[i][j] = 0;
            }else{
                dist[i][j] = 1000000000;
            }
        }
    }

    for(int i = 0; i < M; i++){
        dist[a[i]][b[i]] = min(dist[a[i]][b[i]], c[i]);
        dist[b[i]][a[i]] = min(dist[b[i]][a[i]], c[i]);
    }

    for(int k = 0; k < N; k++){
        for(int i = 0; i < N; i++){
            for(int j = 0; j < N; j++){
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
            }
        }
    }

    int ans = M;

    for(int i = 0; i < M; i++){
        int f = 1;
        for(int j = 0; j < N; j++){
            if(dist[j][a[i]] + c[i] == dist[j][b[i]]){
                f = 0;
            }
        }
        if(f == 0){
            ans--;
        }
    }

    printf("%d\n", ans);

    return 0;
}