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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdint.h> // int64_t

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))

int get_int(void) {
  int num;
  scanf("%d", &num);
  return num;
}

int get_int2(int *a1, int *a2) {
  scanf("%d %d", a1, a2);
  return 0;
}

int get_int3(int *a1, int *a2, int *a3) {
  scanf("%d %d %d", a1, a2, a3);
  return 0;
}

#define NUM_MAX 100

int main(void) {
    int vs, es;
    get_int2(&vs, &es);
    static int graph[NUM_MAX][NUM_MAX];
    int i;
    for(i = 0; i < es; i++) {
        int v1, v2, cost;
        get_int3(&v1, &v2, &cost);
        // 1-indexed -> 0-indexed
        v1--; v2--;
        graph[v1][v2] = graph[v2][v1] = cost;
    }
    static int origin[NUM_MAX][NUM_MAX];
    memcpy(origin, graph, sizeof(int)*NUM_MAX*NUM_MAX);

    int j, k;
    // Warshall floyd
    for(k = 0; k < vs; k++) {
        for(i = 0; i < vs; i++) {
            for(j = 0; j < vs; j++) {
                graph[i][j] = min(graph[i][j], graph[i][k]+graph[k][j]);
            }
        }
    }

    int cnt = 0;
    for(i = 0; i < vs; i++) {
        for(j = i+1; j < vs; j++) {
            if(!origin[i][j]) continue;
            if(origin[i][j] != graph[i][j]) cnt++;
        }
    }
    printf("%d\n", cnt);
    return 0;
}