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 <queue>
#include <vector>
#include <functional>
#include <utility>
#include <algorithm>
#include <set>

using P = std::pair<int, int>;

const int INF = 1e9;

int n, m;
std::vector<std::vector<P>> edges;
std::set<P> used;
int D[110];
std::vector<std::set<P>> us;

inline
P mp(int x, int y) {
    if(x > y) {
        std::swap(x, y);
    }
    return {x, y};
}

void dijkstra(int s) {
    for(int i = 0; i < n; ++i) {
        D[i] = INF;
    }
    us.clear();
    us.resize(n);
    D[s] = 0;
    
    std::priority_queue<P, std::vector<P>, std::greater<P>> que;
    
    que.push({0, s});
    
    while(!que.empty()) {
        auto p = que.top(); que.pop();

        if(p.first > D[p.second]) {
            continue;
        }
        
        for(auto& q : edges[p.second]) {
            if(D[q.first] > D[p.second] + q.second) {
                us[q.first].clear();// = decltype(us[q.first]){};
                us[q.first].insert(mp(p.second, q.first));
                D[q.first] = D[p.second] + q.second;
                que.push({D[q.first], q.first});
            }else if(D[q.first] == D[p.second] + q.second) {
                us[q.first].insert(mp(p.second, q.first));
            }
        }
    }
    
    for(int i = 0; i < n; ++i) {
        for(auto& p : us[i])
            used.insert(p);
    }
}

int main() {
    std::cin >> n >> m;
    
    edges.resize(n);
    
    for(int i = 0; i < m; ++i) {
        int x, y, z;
        std::cin >> x >> y >> z;
        
        edges[--x].push_back({--y, z});
        edges[y].push_back({x, z});
    }
    
    for(int i = 0; i < n; ++i) {
        dijkstra(i);
    }
    
    std::cout << m - used.size() << std::endl;
}