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: Runtime Error)

#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <algorithm>
#include <cmath>
using namespace std;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define MAX3(a, b, c) ((a) > (MAX(b, c)) ? (a) : (MAX(b, c)))
#define MIN3(a, b, c) ((a) < (MIN(b, c)) ? (a) : (MIN(b, c)))

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

int main()
{
    int n, m;
    cin >> n >> m;
    int adj[100][100] = {};
    int usd[100][100] = {};
    pii edge[1000];
    int dist[100];
    rep(i, m) {
        dist[i] = 1 << 8;
        rep(k, m) {
            adj[i][k] = 0;
            usd[i][k] = 0;
        }
    }
    rep(i, m) {
        int a, b, c;
        cin >> a >> b >> c;
        edge[i].first = a - 1;
        edge[i].second = b - 1;
        adj[a - 1][b - 1] = c;
        adj[b - 1][a - 1] = c;
    }

    dist[0] = 0;
    queue<int> q;
    q.push(0);
    while (!q.empty()) {
        int cur = q.front();
        q.pop();
        for (int i = 0; i < n; i++) {
            if (cur == i) continue;
            if (adj[cur][i] > 0) {
                if (dist[i] > adj[cur][i] + dist[cur]) {
                    q.push(i);
                    dist[i] = adj[cur][i] + dist[cur];
                }
            }
        }
    }
    int cnt = 0;
    for (int i = 0; i < m; i++) {
        int x = edge[i].first;
        int y = edge[i].second;
        if (adj[x][y] > abs(dist[x] - dist[y])) {
            cnt++;
        }
    }
    cout << cnt << endl;
    return 0;
}