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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

//答え見たよ！！！
#include<stdio.h>
#define INFINITY 9999999

struct Neighborhood{
  int dist;//結ばれた2つのノードの距離
  int no;//辺の番号
};

int main(){
  int i, j;
  //入力
  int N, M;
  scanf("%d %d", &N, &M);
  int a[M+1], b[M+1], c[M+1];//a[0], b[0], c[0]は使わない
  for(i=1; i<=M; i++)
    scanf("%d %d %d", &a[i], &b[i], &c[i]);
  //入力ここまで
  int dist[N+1][N+1];//最短距離
  //初期値の設定
  for(j=1; j<=N; j++){
    for(i=1; i<=N; i++){
      if(i=j){
        dist[i][j] = 0;
      }else{
        dist[i][j] = INFINITY;
      }
    }
  }
  
  struct Neighborhood nbh[N+1][N+1];
  //頂点が直接結ばれているならばを距離と辺の番号を代入
  for(i=1; i<=M; i++){
    nbh[a[i]][b[i]].dist = c[i];
    nbh[b[i]][a[i]].dist = c[i];
    nbh[a[i]][b[i]].no = i;
    nbh[b[i]][a[i]].no = i;
  }
  
  //各頂点を始点とした最短経路をDijkstra法で求め、
  //それぞれの辺が最短経路に含まれるかをみる
  int s;//スタートとして使用
  int t;//暫定的な起点として使用
  int from[N+1];//点sを起点とする最短距離において、各頂点に対しどの辺から来たかを代入する
  int fix[N+1];//各頂点について値が確定したら1、そうでないなら0
  int temp_min;//最小値を記録
  int min_edge;//最小値をもつ頂点
  int cnt;//N-1回最短距離を確定させるためのカウンター
  int use[M+1];//各辺について最短経路として使われたら1
  for(s=1; s<N; s++){
    for(i=1; i<=N; i++)
      fix[i] = 0;
    fix[s] = 1;
    t = s;
    for(cnt=1; cnt<N; cnt++){
      //このときdist(s,t)は確定していることに注意
      //頂点tに隣り合っていてなおかつ確定していない場所をすべて探索する
      for(i=1; i<=N; i++){
        if(nbh[t][i].dist && !fix[i]){
          dist[s][i] = dist[s][t] + nbh[t][i].dist;
          from[i] = nbh[t][i].no;
        }
      }
      //このとき未確定の頂点のうち距離が最小のものが確定する
      temp_min = 9999;
      for(i=1; i<=N; i++){
        if(!fix[i] && dist[s][i] < temp_min){
          temp_min = dist[s][i];
          min_edge = i;
        }
      }
      fix[min_edge] = 1;
      t = min_edge;
    }
    //頂点s始点のすべての最短距離が出たので使われた経路を記録する
    for(i=1; i<=N; i++)
      use[from[i]] = 1;
  }
  //すべての頂点を始点とした最短距離を探索し終わったので
  int ans = 0;//最短経路に含まれない辺の本数を数え上げる
  for(i=1; i<=M; i++){
    if(!use[i])
      ans++;
  }
  printf("%d\n", ans);
  return 0;
}