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 Python (3.4.3) (judge verdict: Wrong Answer)

n,m=map(int,input().split())
from collections import defaultdict
dist=defaultdict(int)
branch=[]
for k in range(m):
    branch.append(tuple(map(int,input().split())))
branch.sort(key=lambda x:x[2],reverse=True)
ans=0
for k in range(m):
    check=branch.pop()
    a,b,c=check[0],check[1],check[2]
    if dist[(a,b)]==0:
        dist[(a,b)]=c
        dist[(b,a)]=c
        for i in range(1,1+n):
            if i==a or i==b:continue
            if dist[(a,i)]!=0 and dist[(b,i)]==0:
                dist[(b,i)]=c+dist[(a,i)]
                dist[(i,b)]=c+dist[(a,i)]
            if dist[(b,i)]!=0 and dist[(a,i)]==0:
                dist[(a,i)]=c+dist[(b,i)]
                dist[(i,a)]=c+dist[(b,i)]
    elif dist[(a,b)]<=c:
        ans+=1
    elif dist[(a,b)]>c:
        dist[(a,b)]=c
        dist[(b,a)]=c
        for i in range(1,1+n):
            if i==a or i==b:continue
            if dist[(a,i)]!=0 and dist[(b,i)]==0:
                dist[(b,i)]=c+dist[(a,i)]
                dist[(i,b)]=c+dist[(a,i)]
            if dist[(b,i)]!=0 and dist[(a,i)]==0:
                dist[(a,i)]=c+dist[(b,i)]
                dist[(i,a)]=c+dist[(b,i)]
print(ans)