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 <vector>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <utility>
#include <functional>
#include <tuple>
 
using namespace std;
 
#define rep(i,x) for(int i=0;i<x;++i)
#define fst first
#define scd second
 
using pii=pair<int,int>;
using vpii=vector<pii>;
using vi=vector<int>;
using vvi=vector<vi>;
using vs=vector<string>;
 
using data=tuple<int, int, int>;
 
struct edge{
    int to, cost, id;
};
 
vector<edge> G[105];
bool used[1005];
int d[105];
 
void dijkstra(int s)
{
    fill_n(d, 105, 1e9);
    priority_queue<data, vector<data>, greater<data>> q;
 
    d[s]=0;
 
    q.push(data(0, s, 1001));
 
    while (q.size()) {
        data dat = q.top(); q.pop();
 
        int sum, v, eid;
        tie(sum, v, eid) = dat;
 
        if (d[v] < sum) continue;
        used[eid] = true;
 
        for (edge e : G[v]) {
            if (d[e.to] >= d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                q.push(data(d[e.to], e.to, e.id));
            }
        }
    }
}
 
signed main()
{
    int N, M; cin >> N >> M;
 
    rep(i, M) {
        int a, b, c; cin >> a >> b >> c;
        --a, --b;
 
        G[a].push_back(edge{b, c, i});
        G[b].push_back(edge{a, c, i});
    }
 
    rep(i, N) dijkstra(i);
 
    int ans = 0;
 
    rep(i, M) {
        if (!used[i]) ans++;
    }
 
    cout << ans << endl;
}