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>

int N,M,ans;
int gra[101][101];
int pgra[101][101];
int used[101][101];

int min(int x,int y){
	if(x<y)return x;
	return y;
}

void search(int start,int goal){
	int i;
	int cur;
	cur=start;
	while(cur!=goal){
		for(i=1;i<=N;i++)
			if(i!=cur && pgra[cur][i]+gra[i][goal]==gra[cur][goal]){
				if(used[cur][i]==0)ans--;
				used[cur][i]=1;
				used[i][cur]=1;
				cur=i;
				break;
			}
	}
}

int main(){
	
	scanf("%d%d",&N,&M);
	ans=M;
	
	for(int i=1;i<=N;i++)
		for(int j=1;j<=N;j++)if(i!=j)gra[i][j]=9999999;
	
	for(int i=0;i<M;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		gra[a][b]=gra[b][a]=c;
	}
	
	for(int i=1;i<=N;i++)
		for(int j=1;j<=N;j++)pgra[i][j]=gra[i][j];
	
	
	for(int k=1;k<=N;k++)
		for(int i=1;i<=N;i++)
			for(int j=1;j<=N;j++){
				gra[i][j]=min(gra[i][j],gra[i][k]+gra[k][j]);
			}
	
	for(int i=1;i<N;i++)
		for(int j=i+1;j<=N;j++)search(i,j);
			
	printf("%d\n",ans);
	
	return 0;
}