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 Go (1.6) (judge verdict: Runtime Error)

package main
 
import (
  "fmt"
  "math"
)

const inf = math.MaxInt32

type edge struct {
  from int
  to   int
  cost int
}

func main() {
  var n,m int
  fmt.Scan(&n,&m)
  
  edges := make([]edge ,3)
  for i := range edges {
    fmt.Scan(&edges[i].from)
    fmt.Scan(&edges[i].to)
    fmt.Scan(&edges[i].cost)
  }
  
  //ノードi,j間の最短経路を定義
  dist := make([][]int, n)
  for i := 0; i < n; i++ {
    dist[i] = make([]int, n)
    for j := 0; j < n; j++ {
      if i == j {
        dist[i][j] = 0
      } else {
        dist[i][j] = inf
      }
    }
  }
  
  //既知の経路
  for i := 0; i < m; i++ {
    dist[edges[i].from-1][edges[i].to-1] = edges[i].cost
    dist[edges[i].to-1][edges[i].from-1] = edges[i].cost
  }
  
  //スタートとゴールを固定 経由地点を切り替えながら最短経路を更新
  //i:スタート,j:ゴール,k:経由地点
    for i := 0; i < n; i++ {
      for j := 0; j < n; j++ {
        for k := 0; k < n; k++ {
        
        //スタートとゴールが同じ場合は計算不要
        if i == j {
          continue
        }
        //経由した方が低コストの場合、最短経路を更新
        if dist[i][k]+dist[k][j] < dist[i][j] {
          dist[i][j] = dist[j][i]+dist[i][k]
          dist[j][i] = dist[j][i]+dist[i][k]
        }
      }
    }
  }
  
  ans := 0
  
  //最短経路に含まれていない辺をカウント
  for i := 0; i < m; i++ {
    if dist[edges[i].from-1][edges[i].to-1] != edges[i].cost||dist[edges[i].to-1][edges[i].from-1] != edges[i].cost  {
      ans++
    }
  }
  
  fmt.Println(ans)
}