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 "bits/stdc++.h"

#define REP(i,n) for(int i=0;i<n;++i)
#define RREP(i,n) for(int i=n-1;i>=0;--i)
#define FOR(i,m,n) for(int i=m;i<n;++i)
#define RFOR(i,m,n) for(int i=n-1;i>=m;--i)
#define ALL(v) v.begin(),v.end()
#define PB(a) push_back(a)
#define INF 1000000001
#define MOD 1000000007
#define EPS 1e-9

const int dx[8] = { 1,1,0,-1,-1,-1,0,1 };
const int dy[8] = { 0,1,1,1,0,-1,-1,-1 };

using namespace std;


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, m;
	cin >> n >> m;
	vector<vector<pair<int, int>>> v(n, vector<pair<int, int>>(0, { 0,0 }));
	vector<vector<vector<int>>> p(n, vector<vector<int>>(n, vector<int>(1, -1)));
	REP(i, m) {
		int a, b, c;
		
		cin >> a >> b >> c;
		a--;b--;
		v[a].push_back({ b, c });
		v[b].push_back({ a,c });
		p[a][b][0] = b;
		p[b][a][0] = a;
	}
	
	vector<vector<int>> d(n, vector<int>(n, INF));
	REP(i, n) {
		REP(j, v[i].size()) {
			d[i][v[i][j].first] = v[i][j].second;
			d[v[i][j].first][i] = v[i][j].second;
			
		}
	}
	REP(k, n) {
		REP(i, n) {
			REP(j, n) {
				if (d[i][j] > d[i][k] + d[k][j]) {
					d[i][j] = d[i][k] + d[k][j];
					p[i][j].clear();
					REP(s, p[i][k].size())p[i][j].push_back(p[i][k][s]);
					REP(t, p[k][j].size())p[i][j].push_back(p[k][j][t]);
				}
			}
		}
	}
	vector<vector<bool>> used(n, vector<bool>(n, false));

	REP(i, n) {
		REP(j, n) {
			if (p[i][j][0] != -1) {
				int a = i, b;
				REP(k, p[i][j].size()) {
					b = p[i][j][k];
					used[a][b] = true;
					a = b;
				}
			}
		}
	}
	int cnt = 0;
	REP(i, n) {
		REP(j, n) {
			if (!used[i][j]) cnt++;
		}
	}

	cout << (cnt -n*n+2*m)/2<< endl;
	return 0;
}