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<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define sc(x) scanf("%d",&x);
static const int MAX = 100;
static const int INF = 1e9;
int n, m, d[MAX][MAX];
vector<pair<int, P> > v;

void warshall_floyd(){
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            for (int k = 0; k < n; k++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
        }
    }
}

int check(){
    int res = 0;
    bool shortest;
    for (int i = 0; i < m; i++){
        shortest = true;
        for (int j = 0; j < n; j++){
            if (!shortest) break;
            if (d[j][v[i].second.first] + v[i].first \
            == d[j][v[i].second.second]){
                if (shortest) res += 1;
                shortest = false;
            }
        }
    }
    return res;
}

int main(){
    int a, b, c, ans;
    sc(n) sc(m)
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++) d[i][j] = INF;
        d[i][i] = 0;
    }
    for (int i = 0; i < m; i++){
        sc(a) sc(b) sc(c)
        d[a - 1][b - 1] = c;
        d[b - 1][a - 1] = c;
        v.push_back(make_pair(c, P(a - 1, b - 1)));
    }
    warshall_floyd();
    ans = check();
    cout << m - ans << endl;
    return 0;
}