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;

const int MAXN = 1007;

bitset<MAXN> edg;
bitset<MAXN> edgg;

typedef pair<int,int> ii;
typedef pair<ii,int> iii;
typedef vector<iii> vii;

#define fi first
#define se second
	
vii adjList[MAXN];

int dist[MAXN],ans;
priority_queue< ii,vector<ii>,greater<ii> > pq;

int main() {
    int N, E, S;
    cin >> N >> E;

    int u, v, w;
    for (int i = 0; i < E; i++) {
        cin >> u >> v >> w;

        adjList[u].push_back(iii(ii(w,i),v));
    }

for(int S=1;S<=N;S++)
{
    memset(dist,-1,sizeof(dist));
    dist[S] = 0; pq.push(ii(0,S));
    while (!pq.empty()) {
        ii front = pq.top(); pq.pop();
        int u = front.se;
        int d = front.fi;

        if (d > dist[u]) continue;

        for (int i = 0; i < adjList[u].size(); i++){
            int vv = adjList[u][i].se;
            int vd = adjList[u][i].fi.fi;

            if (dist[vv] == -1 || dist[vv] > dist[u] + vd){
            	dist[vv] = dist[u] + vd, pq.push(ii(dist[vv],vv));
            	edg[adjList[u][i].fi.se]=1;
            	edg[adjList[u][vv].fi.se]=0;
            }
        }
    }
    
    for (int i=1; i<=N; i++){
    	if(edg[i]==1) edgg[i]=1;
    	edg[i]=0;
    }
}

    for (int i=1; i<=N; i++)
        if (edgg[i] == 0) ans++;
        
    cout << ans << endl;
  
    return 0;
}