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 (2.7.6) (judge verdict: Wrong Answer)

#!/usr/bin/python

import sys

def main(argv):
    line = sys.stdin.readline().rstrip("\n")

    while line:
        n, m = line.split()
        n = int(n)
        m = int(m)

        edges = []
        path = [[[] for i in xrange(n)] for i in xrange(n)]
        distance = [[0 for i in xrange(n)] for i in xrange(n)]
        for i in xrange(m):
            a, b, c = raw_input().split()
            a = int(a) - 1
            b = int(b) - 1
            c = int(c)
            distance[a][b] = c
            distance[b][a] = c
            edges.append((a,b))

        for i in xrange(n):
            for j in xrange(n):
                for k in xrange(n):
                    candidate = distance[i][k] + distance[k][j]
                    if candidate < distance[i][j]:
                        distance[i][j] = candidate
                        path[i][j].append(k)

        counter = 0
        for (x, y) in edges:
            if path[x][y]:
                counter += 1
        print counter

        line = sys.stdin.readline().rstrip("\n")

if __name__ == "__main__":
    main(sys.argv)