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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

typedef long long int int64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))

#define POS(i,j) ((i)*n+(j))

void run(void){
  int n,m;
  scanf("%d%d",&n,&m);
  const int inf=100000+1;
  int cost[10000];
  int i,j,k;
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      cost[POS(i,j)]=i==j?0:inf;
    }
  }
  for(i=0;i<m;i++){
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    a--;b--;
    cost[POS(a,b)]=c;
    cost[POS(b,a)]=c;
  }
  int dp[10000];
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      dp[POS(i,j)]=cost[POS(i,j)];
    }
  }
  for(k=0;k<n;k++){
    for(i=0;i<n;i++){
      for(j=0;j<n;j++){
	if(dp[POS(i,j)]>dp[POS(i,k)]+dp[POS(k,j)]) dp[POS(i,j)]=dp[POS(i,k)]+dp[POS(k,j)];
      }
    }
  }
  int cnt=0;
  for(i=0;i<n;i++){
    for(j=i+1;j<n;j++){
      if(cost[POS(i,j)]!=inf && dp[POS(i,j)]<cost[POS(i,j)]) cnt++;
    }
  }
  printf("%d\n",cnt);
}

int main(void){
  run();
  return 0;
}