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: Wrong Answer)

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

int main(){
	int N,M;
	int a,b,c,d[100][100]={},e[100][100]={};
	int i,j,k;
	int z,count;
	int min;
	
	scanf("%d %d",&N,&M);
	for(i=0;i<M;i++){
		scanf("%d %d %d",&a,&b,&c);
		d[a][b]=c;
		d[b][a]=c;
	}
	
	for(i=1;i<=N;i++){
		for(j=1;j<i;j++){
			z=0;
			for(k=1;k<=N;k++){
				if(d[i][j]!=0){
					min=d[i][j];
				}else{
					min=d[j][i];
				}
				e[i][j]=1;
				e[j][i]=1;
				if(d[i][k]+d[k][j]<min && i!=j){
					min=d[i][k]+d[k][j];
					e[i][k]=1;
					e[k][i]=1;
					e[k][j]=1;
					e[j][k]=1;
					if(z!=0){
						e[i][z]=0;
						e[z][i]=0;
						e[z][j]=0;
						e[j][z]=0;
					}else{
										e[i][j]=0;
				e[j][i]=0;
					}
					z=k;
				}
			}
		}
		for(j=i+1;j<=N;j++){
			z=0;
			for(k=1;k<=N;k++){
				if(d[i][j]!=0){
					min=d[i][j];
				}else{
					min=d[j][i];
				}
				e[i][j]=1;
				e[j][i]=1;
				if(d[i][k]+d[k][j]<min && i!=j){
					min=d[i][k]+d[k][j];
					e[i][k]=1;
					e[k][i]=1;
					e[k][j]=1;
					e[j][k]=1;
					if(z!=0){
						e[i][z]=0;
						e[z][i]=0;
						e[z][j]=0;
						e[j][z]=0;
					}else{
										e[i][j]=0;
				e[j][i]=0;
					}
					z=k;
				}
			}
		}
	}
	for(i=1;i<=N;i++){
		for(j=1;j<=N;j++){
			if(e[i][j]==1){
				count++;
			}
		}
	}
	
	
	printf("%d\n",M-(count/2));
	
	
	
	return 0;
}