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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	
	static int INF = 10000000;

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int[][] d = new int[N+1][N+1];
		int[][] e = new int[N+1][N+1];
		for(int i=1; i<=N; i++) {
			for(int j=i+1; j<=N; j++) {
				d[i][j] = INF;
			}
		}
		for(int i=0; i<M; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int c = sc.nextInt();
			d[a][b] = c;
			e[a][b] = c;
		}
		for(int k=1; k<=N; k++) {
			for(int i=1; i<=N; i++) {
				for(int j=i+1; j<=N; j++) {
					d[i][j] = Math.min(d[i][j], d[i][k]+d[k][j]);
				}
			}
		}
		int ans = 0;
		for(int i=1; i<=N; i++) {
			for(int j=i+1; j<=N; j++) {
				if(e[i][j] != 0 && d[i][j] < e[i][j]) ans++;
			}
		}
		System.out.println(ans);
	}
	
	// print
	static void print(String s) {
		System.out.println(s);
	}
	
	// union find lib
	// usage:
	// 最初にinitを呼ぶ
	// root: 直接は呼ばないで
	// unite: まとめる
	// same: グループ判定
	static void init(int par[], int N) {
		for(int i=0; i<N; i++) {
			par[i] = i;
		}
	}
	
	static int root(int x, int [] par) {
		if(par[x] == x) {
			return x;
		} else {
			return (par[x] = root(par[x], par));
		}
	}
	
	static boolean same(int x, int y, int[] par) {
		return root(x, par) == root(y, par);
	}
	
	static void unite(int x, int y, int[] par) {
		x = root(x, par);
		y = root(y, par);
		if(x == y) return;
		par[x] = y;
	}
	
	// end union find lib
	
	// number lib
	
	static long lcm(long a, long b) {
		return a*(b/gcd(a, b));
	}
	
	static long gcd(long a, long b) {
		long ta = Math.max(a, b);
		long tb = Math.min(a, b);
		long tmp;
		while((tmp = ta%tb) != 0) {
			ta = tb;
			tb = tmp;
		}
		return tb;
	}
	
	static long modcomb(long n, long k, long mod) {
		if(k==1) {
			return n;
		}
		
		long ans = 1;
		for(long i=n; i>=n-k+1; i--) {
			ans = (ans * i)%mod;
		}
		for(long i=k; 0<i; i--) {
			ans = (ans * modpow(i, mod-2, mod)) % mod;
		}
		return ans;
	}
	
	static long modpow(long a, long b, long mod) {
		if(b==0) return 1;
		if(b%2==0) {
			long d = modpow(a, b/2, mod);
			return (d*d)%mod;
		} else {
			return (a*modpow(a, b-1, mod))%mod;
		}
	}
	
	static int disit(long a, long d) {
		int count = 0;
		while(a!=0) {
			a = a/d;
			count++;
		}
		return count;
	}
	
	// end number lib
}