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<math.h>
#include<stdlib.h>
#include<limits.h>

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define revrep(i,begin,end) for(int i=begin; i>end; i--)
#define lld long long int

int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    int c[m], f[m];
    int wf[n+1][n+1];
    int len[n+1][n+1];
    int dir[n+1][n+1][n];
    rep(i, 0, n+1){
        rep(j, 0, n+1){
            len[i][j] = 0;
            wf[i][j] = 1000000;
            if(i==j){
                wf[i][j] = 0;
            }
        }
    }
    rep(i, 0, m){
        int a, b;
        scanf("%d %d %d", &a, &b, &c[i]);
        f[m] = 1;
        wf[a][b] = c[i];
        wf[b][a] = c[i];
        dir[a][b][len[a][b]++] = i;
        dir[b][a][len[b][a]++] = i;
    }
    rep(k, 1, n+1){
        rep(i, 1, n+1){
            rep(j, 1, n+1){
                if(wf[i][j] > wf[i][k] + wf[k][j]){
                    wf[i][j] = wf[i][k] + wf[k][j];
                    int x = len[i][k] + len[k][j];
                    rep(l, 0, len[i][k]){
                        dir[i][j][l] = dir[i][k][l];
                    }
                    rep(l, 0, len[k][j]){
                        dir[i][j][len[i][k]+l] = dir[k][j][l];
                    }
                    len[i][j] = x;
                }
            }
        }
    }
    rep(i, 1, n+1){
        rep(j, 1, n+1){
            rep(k, 0, len[i][j]){
                f[dir[i][j][k]] = 0;
            }
        }
    }
    int ans = 0;
    rep(i, 0, m){
        if(f[i]){
            ans++;
        }
    }
    printf("%d\n", ans);

    return 0;
}