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>
using namespace std;
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
typedef long long int ll;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define MOD (1000000007)
#define ALL(v) (v).begin(), (v).end()
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

//-------------------------------------

struct edge
{
    int from;
    int to;
    int cost;
};

struct UnionFind
{
    vector<int> par;

    UnionFind(int n) : par(n, -1) {}
    void init(int n) { par.assign(n, -1); }

    int root(int x)
    {
        if (par[x] < 0)
            return x;
        else
            return par[x] = root(par[x]);
    }

    bool issame(int x, int y)
    {
        return root(x) == root(y);
    }

    bool merge(int x, int y)
    {
        x = root(x);
        y = root(y);
        if (x == y)
            return false;
        if (par[x] > par[y])
            swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    int size(int x)
    {
        return -par[root(x)];
    }
};

int kruskal(vector<edge> &graph, int V)
{
    sort(ALL(graph), [](const edge &a, const edge &b) { return a.cost < b.cost; });
    UnionFind uf(V);

    int res = 0;
    for (int v = 0; v < (int)graph.size(); v++)
    {
        edge e = graph[v];
        if (!uf.issame(e.from, e.to))
        {
            uf.merge(e.from, e.to);
            res++;
        }
    }
    return res;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    vector<edge> graph(m);
    for (int i = 0; i < m; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        a--;
        b--;
        graph[i] = edge{a, b, c};
    }
    int ans = kruskal(graph, n);
    cout << m - ans << endl;
}