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: Accepted)

#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
ll W[110][110];
ll init[110][110];
ll counts[110][110];
struct edge {
	int v1, v2;
	int value;
}edges[1024];
//#define LOCAL
#define INF 0x3f3f3f3f;
void run() {
	int N, M;
	while(cin >> N >> M) {
		memset(init, 0x3f, sizeof(init));
		memset(W, 0x3f, sizeof(W));
		memset(counts, 0, sizeof(counts));
		int v1, v2, value;
		for(int i = 0; i < M; i++) {
			cin >> v1 >> v2 >> value;
			init[v1][v2] = init[v2][v1] = W[v1][v2] = W[v2][v1] = value;
			edges[i].v1 = v1;
			edges[i].v2 = v2;
			edges[i].value = value;
		}
		#ifdef LOCAL 
		for(int i = 1; i <= N; i++) {
			for(int j = 1; j <= N; j++) {
				cout << W[i][j] << " ";
			}
			cout << endl;
		}
		#endif
		for(int l = 1; l <= N; ++l) {
			for(int i = 1; i <= N; i++) {
				for(int j = 1; j <= N; j++) {
					W[i][j] = min(W[i][j], W[i][l] + W[l][j]);
				}
			}
		}
		#ifdef LOCAL 
		for(int i = 1; i <= N; i++) {
			for(int j = 1; j <= N; j++) {
				cout << init[i][j] << " ";
			}
			cout << endl;
		}
		#endif
		ll ans = 0;
		for(int i = 0; i < M; i++) {
			#ifdef LOCAL 
			 cout << init[edges[i].v1][edges[i].v2] << " " << edges[i].value << endl;
			#endif
			if(W[edges[i].v1][edges[i].v2] < edges[i].value)
				ans++;
		}
		cout << ans << endl;

	}
}
int main(int argc, char *argv[]) {
	run();
}