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 "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "math.h"
#include "utility"
#include "string"
#include "map"
#include "unordered_map"
#include "iomanip"
#include "random"

using namespace std;
const long long int MOD = 1000000007;

long long int power(long long int x, long long int n, long long int M) {
	long long int tmp = 1;

	if (n > 0) {
		tmp = power(x, n / 2, M);
		if (n % 2 == 0) tmp = (tmp*tmp) % M;
		else tmp = (((tmp*tmp) % M)*x) % M;
	}
	return tmp;
}

long long int N, M, K, Q, W, H, L, R;
long long int ans;



int main() {
	ios::sync_with_stdio(false);
	int num[101][101] = {};
	bool used[101][101] = {};
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		num[a][b] = c;
		num[b][a] = c;
	}
	priority_queue<pair<int, int>>PQ;
	for (int i = 1; i <= N; i++) {
		vector<int>dis(N + 1, INT_MAX);
		dis[i] = 0;
		PQ.push({ 0,i });
		while (!PQ.empty()) {
			int c = PQ.top().first;
			int cn = PQ.top().second;
			PQ.pop();
			if (dis[cn] < c)continue;
			dis[cn] = c;
			for (int j = 1; j <= N; j++) {
				if (num[cn][j]) {
					if (dis[j] > dis[i] + num[cn][j]) {
						dis[j] = dis[i] + num[cn][j];
						PQ.push({ dis[j],j });
					}
				}
			}
		}
		for (int j = i + 1; j <= N; j++) {
			queue<int>QQ;
			QQ.push(j);
			while (!QQ.empty()) {
				int c = QQ.front();
				for (int k = 1; k <= N; k++) {
					if (num[c][k]) {
						if (dis[c] == dis[k] + num[k][c]) {
							QQ.push(k);
							used[c][k] = true;
							used[k][c] = true;
						}
					}
				}
				QQ.pop();
			}
		}
	}
	for (int i = 1; i <= N; i++) {
		for (int j = i + 1; j <= N; j++) {
			if (num[i][j] && !used[i][j])ans++;
		}
	}
	cout << ans << endl;
	return 0;
}