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<stdio.h>
#include<iostream>
using namespace std;

int N,M;
int d[110][110];
bool e[110][110] ;
int INF = 2147483647;
int main (){
    cin >> N >> M;

    for(int i=0; i<N; i++){
        for(int j=0; j<N; j++){
            e[i][j] = false;
            if( i == j ) {
                d[i][j] = 0;
                e[i][j] = true;
            }
            else d[i][j] = INF;
        }
    }

    for(int i=0; i<M; i++){
        int a,b,c;
        cin >> a >> b >> c;
        a--; b--;
        d[a][b] = d[b][a] = c;
    }

    for(int k=0; k<N; k++){
        for(int i=0; i<N; i++){
            for(int j=0; j<N; j++){
                if( d[i][j] < d[i][k] + d[k][j] ) {
                    e[i][j] = true;
                } else if ( d[i][j] > d[i][k] + d[k][j] ) {
                    e[i][k] = true;
                    e[k][j] = true;
                }
            }
        }
    }

    int cnt = 0;
    for(int i=0; i<N; i++){
        for(int j=0; j<N; j++){
            if( e[i][j] == false ) cnt++;
        }
    }
    cout << cnt / 2 << endl;
    return 0;
}