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)

from copy import deepcopy
N,M=list(map(int, input().strip().split()))
abc={}
used={}
for _ in range(M):
    a,b,c=(list(map(int, input().strip().split())))
    abc[a,b]=c
    used[a,b]=False
spth=[[float("inf")]*(N+1) for _ in range(N+1)]
for i in range(1,N+1):
    for j in range(1,N+1):
        if i==j:
            spth[i][j]=0
        if (i,j) in abc:
            spth[i][j]=abc[i,j]
        elif (j,i) in abc:
            spth[i][j]=abc[j,i]
for i in range(1,N+1):
    for j in range(1,N+1):
        for k in range(1,N+1):
            d=spth[i][k]+spth[k][j]
            if spth[i][j] > d:
                spth[i][j]=d

for s in range(1,N+1):
    for ab,c in abc.items():
        if spth[s][ab[1]]==spth[s][ab[0]]+c:
            used[ab]=True
            break
print(sum(not i for i in used.values()))