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 cost[105][105];
int d[105];
int used[105];
int d2[105][105];

int N,M;
int a[1005];
int b[1005];
int c[1005];

void dijkstra(int s){
	int i;

	for(i=0;i<N;i++){
		d[i]=1<<20;
		used[i]=0;
	}
	d[s]=0;

	while(1){
		int v=-1;

		for(i=0;i<N;i++){
			if(used[i]==0&&(v==-1 || d[i]<d[v])){v=i;}
		}
		if(v==-1){break;}

		used[v]=1;

		for(i=0;i<N;i++){
			if(d[i]>d[v]+cost[v][i]){d[i]=d[v]+cost[v][i];}
		}

	}

}

void warshall_floyd(){

	int i,j,k;

	for(i=0;i<N;i++){
		for(j=0;j<N;j++){
			d2[i][j]=cost[i][j];
		}
	}

	for(k=0;k<N;k++){
		for(i=0;i<N;i++){
			for(j=0;j<N;j++){
				if(d2[i][j]>d2[i][k]+d2[k][j]){
					d2[i][j]=d2[i][k]+d2[k][j];
				}
			}
		}
	}

}

int main(){

	scanf("%d %d",&N,&M);

	int i,j;
	for(i=0;i<M;i++){
		scanf("%d %d %d",&a[i],&b[i],&c[i]);
		cost[a[i]-1][b[i]-1]=c[i];
		cost[b[i]-1][a[i]-1]=c[i];
	}

	for(i=0;i<N;i++){
		for(j=0;j<N;j++){
			if(i==j){cost[i][j]=0;}
			else{
				if(cost[i][j]==0){cost[i][j]=1<<20;}
			}
		}
	}

	warshall_floyd();

	int ans=0;

	for(i=0;i<M;i++){
		cost[a[i]-1][b[i]-1]=1<<20;
		cost[b[i]-1][a[i]-1]=1<<20;
		dijkstra(a[i]-1);
		int flag=1;
		for(j=0;j<N;j++){
			if((d2[a[i]-1][j]!=d[j]) || d[j]==1<<20){
				flag=0;
				break;
			}
		}
		ans+=flag;
		cost[a[i]-1][b[i]-1]=c[i];
		cost[b[i]-1][a[i]-1]=c[i];
	}

	printf("%d\n",ans);

	return 0;
}