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: Accepted)

#include<stdio.h>
#define X 1000000000

int minim(int a,int b){if(a > b){return b;}return a;}

int main(void){
	int dp[128][128]={0};
	int n,m,a[1024],b[1024],c[1024];
	int i,j,k,r=0;
	scanf("%d%d",&n,&m);
	for(i = 1;i <= m;i++){scanf("%d%d%d",&a[i],&b[i],&c[i]);}
	for(i = 0;i <= 127;i++){
		for(j = 0;j <= 127;j++){dp[i][j] = X;if(i == j){dp[i][j]=0;}}
	}
	for(k = 1;k <= m;k++){
		dp[(a[k])][(b[k])] = c[k];
		dp[(b[k])][(a[k])] = c[k];
	}
	for(k = 1;k <= n;k++){
		for(i = 1;i <= n;i++){
			for(j = 1;j <= n;j++){dp[i][j] = minim(dp[i][j],dp[i][k]+dp[k][j]);}
		}
	}
	for(k = 1;k <= m;k++){
		if(dp[(a[k])][(b[k])] < c[k]){r++;}
	}
	printf("%d\n",r);
	return 0;
}