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;

struct UnionFind {
	vector<int> par;
	UnionFind(int n) : par(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);
		par[x] += par[y];
		par[y] = x;
		return true;
	}
	int size(int x) {
		return -par[root(x)];
	}
};

struct Edge {
	int a, b, cost;
	bool operator<(const Edge& o) const {
		return cost < o.cost;
	}
};

struct Graph {
	int n;
	vector<Edge> es;

	int kruskal() {
		sort(es.begin(), es.end());
		UnionFind uf(n);
		int res = 0;
		for (int i = 0; i < es.size(); i++)
		{
			Edge& e = es[i];
			if(uf.issame(e.a, e.b)) res++;
			else uf.merge(e.a, e.b);
		}
		return res;
	}
};

Graph input_graph() {
	Graph g;
	int m;
	cin >> g.n >> m;
	for (int i = 0; i < m; i++)
	{
		Edge e;
		cin >> e.a >> e.b >> e.cost;
		e.a--; e.b--;
		g.es.push_back(e);
	}
	return g;
}

int main() {
	Graph g = input_graph();
	cout << g.kruskal() << endl;
    return 0;
}