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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
long long n,m;
std::vector<std::vector<std::pair<long long,long long>>> hen; // ver : cost
std::vector<std::vector<long long>> seen; //辺i-jの使用回数
std::vector<long long> past;

void dijkstra(long long start){
    std::vector<long long> dist(n,10000000000000000);
    using T = std::pair<long long, long long>;
    std::priority_queue<T, std::vector<T>, std::greater<T>> que;
    que.push({0,start});
    dist[start] = 0;
    while(que.size()){
        auto cur = que.top();
        que.pop();
        for(auto i : hen[cur.second]){

            if(dist[cur.second] + i.second < dist[i.first]){
                dist[i.first] = cur.first + i.second;
                seen[i.first][past[i.first]]--;
                seen[past[i.first]][i.first]--;
                seen[i.first][cur.second]++;
                seen[cur.second][i.first]++;
                que.push({dist[i.first],i.first});
                past[i.first] = cur.second;
            }
        }
    }
}


int main()
{
    std::cin >> n >> m;
    hen.resize(n);
    seen.resize(n + 10,std::vector<long long>(n + 10,-1000000000));
    past.resize(n, n + 5);
    for(long long i = 0; i < m; i++){
        long long a,b,c;
        std::cin >> a >> b >> c;
        a--; b--;
        hen[a].push_back({b,c});
        hen[b].push_back({a,c});
        seen[a][b] = 0;
        seen[b][a] = 0;
    }


    for(long long i = 0; i < n; i++)dijkstra(i);
    long long cnt = 0;


    for(long long i = 0; i < n; i++){
        for(long long j = i + 1; j < n; j++){
            if(seen[i][j] == 0)cnt++;
        }
    }


    std::cout << cnt << std::endl;

}