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: Time Limit Exceeded)

# cook your dish here
def dijkstra_back(s,t,n,w,cost):
    #sからtへの最短経路の経路復元
    prev = [s] * n #最短経路の直前の頂点
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        for i in range(n):
            if (not used[i]) and (v == -1):
               v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
               break
        used[v] = True
        
        for i in range(n):
            if d[i] > d[v] + cost[v][i]: 
                d[i] = d[v] + cost[v][i]
                prev[i] = v
        
    path = [t]
    while prev[t] != s:
        path.append(prev[t])
        prev[t] = prev[prev[t]]
    path.append(s)
    path = path[::-1]
    return path

################################
n,w = map(int,input().split()) #n:頂点数　w:辺の数

cost = [[float("inf") for i in range(n)] for i in range(n)] 
#cost[u][v] : 辺uvのコスト(存在しないときはinf この場合は10**10)
for i in range(w):
    x,y,z = map(int,input().split())
    cost[x-1][y-1] = z
    cost[y-1][x-1] = z

check = [[0 for i in range(n)] for j in range(n)]

for i in range(n):
    for j in range(i+1,n):
        tmp = dijkstra_back(i,j,n,w,cost)
        #print(tmp)
        for k in range(len(tmp)-1):
            #print(tmp[k],tmp[k+1])
            check[min(tmp[k],tmp[k+1])][max(tmp[k],tmp[k+1])] = 1

#print(check)

#exit()

print(w - sum(map(sum, check)))

#print(dijkstra_back(0,1,n,w,cost))