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<algorithm>
#include<vector>
#include<queue>
#include<string>
#include<iomanip>
#include<sstream>
#include<map>
#include<cmath>
using namespace std;
#define rep(i,n) for(int i = 0;i < n;i++)
#define req(i,n) for(int i = 1;i <=n;i++)
#define rrep(i,n) for(int i = n -1;i >= 0;i--)
#define ALL(a) a.begin(),a.end()
typedef long long ll;
typedef long double ld;
const ll MOD =  1000000007;
const ll INF =  1LL << 60;
int n, m, x, y;ll z = 0, w, h, sum = 0, ans = 0;
template<typename A, size_t N, typename T>
void Fill(A(&array)[N], const T & val) {
    std::fill((T*)array, (T*)(array + N), val);
}
struct Corr {
    ll x;
    ll y;
    ll z;
};
int main(void) {
    cin >> n >> m;
    ll dp[110][110]; vector<Corr> a(m);
    Fill(dp, INF);
    rep(i, m) {
        cin >> x >> y >> z;
        x--; y--;
        dp[x][y] = z;
        dp[y][x] = z;
        a[i] = { x,y,z };
    }
    rep(i, n) dp[i][i] = 0;
    rep(i, n) {
        rep(j, n) {
            rep(k, n) {
                dp[j][k] = min(dp[j][k], dp[j][i] + dp[i][k]);
            }
        }
    }
    rep(i, m) {
        bool f = 0;
        rep(j, n) {
            if (dp[j][a[i].x] + a[i].z == dp[j][a[i].y]) {
                f = 1; break;
            }
            if (dp[j][a[i].y] + a[i].z == dp[j][a[i].x]) {
                f = 1; break;
            }
        }
        if (!f) ans++;
    }cout << ans << endl;
}