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 <fstream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <numeric>
#include <functional>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <list>

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

using namespace std;
using ll = long long;
using P = pair<ll, ll>;
using TP = tuple<ll, ll, ll>;
ll MOD = 1e9+7;
const int INF = 100000000;
const ll LINF = (ll)1e18;

using Graph = vector<vector<P>>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int n,m;
    cin >> n >> m;
    
    int a[1000],b[1000],c[1000],dist[100][100];
    
    for(int i = 0; i < m; ++i){
        cin >> a[i] >> b[i] >> c[i];
        a[i]--,b[i]--;
    }
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            if(i==j){
                dist[i][j]=0;
            }
            else
            {
                dist[i][j]=INF;
            }
        }
    }
    for(int i = 0; i < m; ++i){
        dist[a[i]][b[i]]=min(dist[a[i]][b[i]],c[i]);
        dist[b[i]][a[i]]=min(dist[b[i]][a[i]],c[i]);
    }
    for(int k = 0; k < n; ++k){
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
            }
        }
    }
    int ans=m;
    for(int i = 0; i < m; ++i){
        bool shortest=false;
        for(int j = 0; j < n; ++j){
            if(dist[j][a[i]]+c[i]==dist[j][b[i]]) shortest=true;
        }
        if(shortest==true){
            ans=ans-1;
        }
    }
    cout << ans << endl;
    return 0;
}