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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); ++i)

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 1e5 + 5;
const int INF = 1001001001;
const int MOD = 1e9 + 7;

class UnionFindTree{
    private:
        vector<int> par, rank;
    
    public:
        UnionFindTree(int n) {
            par = rank = vector<int>(n);

            for (int i = 0; i < n; ++i) {
                par[i] = i;
                rank[i] = 0;
            }
        }

        int find(int x) {
            if (par[x] == x) {
                return x;
            } else {
                return par[x] = find(par[x]);
            }
        }

        void unite(int x, int y) {
            x = find(x);
            y = find(y);
            if (x == y) return;

            if (rank[x] < rank[y]) {
                par[x] = y;
            } else {
                par[y] = x;
                if (rank[x] == rank[y]) rank[x]++;
            }
        }

        bool same(int x, int y) {
            return find(x) == find(y);
        }
};

int main(){
    int N, M;
    cin >> N >> M;

    priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<tuple<int, int, int>>> q;
    rep(i, M) {
        int a, b, c;
        cin >> a >> b >> c;
        a--, b--;
        if (a > b) swap(a, b);
        q.emplace(c, a, b);
    }

    int ans = 0;
    UnionFindTree uf(N);
    while (!q.empty()) {
        auto t = q.top(); q.pop();
        int a = get<1>(t), b = get<2>(t);

        if (uf.same(a, b)) ans++;
        else uf.unite(a, b);
    }

    cout << ans << endl;
}