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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
typedef long long ll;
const int inf = 1e8;
struct edge{int to, cost;};
typedef pair<int, int> P;
typedef pair<int, P> PP;

int N, M;
vector<edge> G[111];
int d[111];

int flag[111][111];
int ans;

void dijkstra(int s){
  priority_queue<PP, vector<PP>, greater<PP> >que;
  FOR(i,1,N+1) d[i] = inf;
  d[s] = 0;
  que.push(PP(0, P(s, s)));

  while(!que.empty()){
    PP pp = que.top(); que.pop();
    int v = pp.second.second;
    if(d[v] < pp.first) continue;
    // 来た頂点u
    int u = pp.second.first;
    if(!flag[u][v] && !flag[v][u] && v!=u){
      ans--;
      flag[u][v] = true;
    }
    for(auto &e : G[v]){
      if(d[e.to] > d[v] + e.cost){
        d[e.to] = d[v] + e.cost;
        que.push(PP(d[e.to], P(v, e.to)));
      }
    }
  }
}

int main(){
  cin >> N >> M;
  FOR(i,0,M){
    int a, b, c;
    cin >> a >> b >> c;
    edge e1;
    e1.to = b;
    e1.cost = c;
    edge e2;
    e2.to = a;
    e2.cost = c;
    G[a].push_back(e1);
    G[b].push_back(e2);
  }
  FOR(i,0,N+1) FOR(j,0,N+1) flag[i][j] = false;
  ans = M;
  FOR(i,1,N+1) dijkstra(i);
  cout << ans << endl;
  return 0;
}