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 INF 1000000000
#define MOD 1000000007
#define MAXN 105
#define lint long long
using namespace std;

struct ii {
	int a, b, c;
	bool operator > (ii o) const {
		return tie(a, b, c) > tie(o.a, o.b, o.c);
	}
};

struct iii {
	int a, b, c, d, e;
	bool operator > (iii o) const {
		return tie(a, b, c, d, e) > tie(o.a, o.b, o.c, o.d, o.e);
	}
};

int TC, N, M, K, Q, x, y, z, cnt, mnm, mks, ptr, ans;
priority_queue<iii, vector<iii>, greater<iii> > pq;
vector<ii> edge[MAXN];
int dist[MAXN][MAXN];

int mod(lint n) { return (n%MOD + MOD)%MOD; }

int main() {
	#ifdef FAIRUZI10
	freopen("/home/fairuzi/Documents/CP/input.in", "r", stdin);
	#endif
	ios::sync_with_stdio(0);
	
	cin>>N>>M;
	edge[0].push_back({-1, -1, -1});
	for (int i = 0; i < M; i++) {
		cin>>x>>y>>z;
		edge[x].push_back({z, y, 0});
		edge[y].push_back({z, x, 0});
	}
	
	for (int i = 1; i <= N; i++) {
		for (int j = 1; j <= N; j++) {
			dist[i][j] = INF;
		}
	}
	for (int i = 1; i <= N; i++) {
		//~ dist, awal, now, prev, idxVec
		pq.push({0, i, i, 0, 0});
		while (!pq.empty()) {
			iii now = pq.top(); pq.pop();
			if (now.a > dist[now.b][now.c]) continue;
			edge[now.d][now.e].c = 1;
			for (int i = 0; i < edge[now.c].size(); i++) {
				int cost = now.a+edge[now.c][i].a;
				if (cost <= dist[now.b][edge[now.c][i].b]) {
					dist[now.b][edge[now.c][i].b] = cost;
					pq.push({cost, now.b, edge[now.c][i].b, now.c, i});
				}
			}
		}
	}
	
	ans = 0;
	for (int i = 1; i <= N; i++) {
		for (ii j: edge[i]) {
			if (!j.c) ans++;
		}
	}
	ans>>=1;
	cout<<ans<<'\n';
	
	return 0;
}