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++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <fstream>
#include <string>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <vector>
#include <algorithm>
#include <numeric>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <functional>
#include <cctype>
#include <list>
#include <limits>
//#include <boost/multiprecision/cpp_int.hpp>

#define BIT(a) (1 << (a))
#define EPS (1e-10)

using namespace std;
//using namespace boost::multiprecision;

const long long MOD = 1000000007;
const int COUNTER_CLOCKWISE = 1;
const int CLOCKWISE = -1;
const int ONLINE_BACK = 2;
const int ONLINE_FRONT = -2;
const int ON_SEGMENT = 0;

class DisjointSet{
    public:
        vector<int> rank, p;

        DisjointSet(){}
        DisjointSet(int size){
            rank.resize(size, 0);
            p.resize(size, 0);
            for (int i = 0; i < size; i++) makeSet(i);
        }

        void makeSet(int x){
            p[x] = x;
            rank[x] = 0;
        }

        bool same(int x, int y){
            return findSet(x) == findSet(y);
        }

        void unite(int x, int y){
            link(findSet(x), findSet(y));
        }

        void link(int x, int y){
            if (rank[x] > rank[y]){
                p[y] = x;
            }else{
                p[x] = y;
                if (rank[x] == rank[y]){
                    rank[y]++;
                }
            }
        }

        int findSet(int x){
            if (x != p[x]){
                p[x] = findSet(p[x]);
            }
            return p[x];
        }
};

class Point{
    public:
    double x, y;

    Point(double x = 0, double y = 0): x(x), y(y) {}

    Point operator+(Point p){return Point(x + p.x, y + p.y);}
    Point operator-(Point p){return Point(x - p.x, y - p.y);}
    Point operator*(double a){return Point(a * x, a * y);}
    Point operator/(double a){return Point(x / a, y / a);}

    bool operator < (const Point &p) const {
        return y != p.y ? y < p.y : x < p.x;
    }

    double norm(){return x * x + y * y;}
};
typedef Point Vector;
typedef vector<Vector> Polygon;

double cross(Vector a, Vector b){
    return a.x*b.y - a.y*b.x;
}

double dot(Vector a, Vector b){
    return a.x * b.x + a.y * b.y;
}

int ccw(Point p0, Point p1, Point p2){
    Vector a = p1 - p0;
    Vector b = p2 - p0;
    if (cross(a, b) > EPS)return COUNTER_CLOCKWISE;
    if (cross(a, b) > -EPS)return CLOCKWISE;
    if (dot(a, b) < -EPS)return ONLINE_BACK;
    if (a.norm() < b.norm())return ONLINE_FRONT;

    return ON_SEGMENT;
}

Polygon andrewScan(Polygon s){
    Polygon u, l;
    if (s.size() < 3) return s;
    sort(s.begin(), s.end());
    u.push_back(s[0]);
    u.push_back(s[1]);
    l.push_back(s[s.size()-1]);
    l.push_back(s[s.size()-2]);

    for (size_t i = 2; i < s.size(); i++){
        for (size_t n = u.size(); n >= 2 && ccw(u[n-2], u[n-1], s[i]) == COUNTER_CLOCKWISE; n--){
            u.pop_back();
        }
        u.push_back(s[i]);
    }

    for (int i = (int)s.size() - 3; i >= 0; i--){
        for (size_t n = l.size(); n >= 2 && ccw(l[n-2], l[n-1], s[i]) == COUNTER_CLOCKWISE; n--){
            l.pop_back();
        }
        l.push_back(s[i]);
    }

    reverse(l.begin(), l.end());
    for (size_t i = u.size() - 2; i >= 1; i--) l.push_back(u[i]);

    return l;
}

long long mod_pow(long long x, long long n){
    long long res = 1;
    for(int i = 0;i < 60; i++){
        if(n >> i & 1) res = res * x % MOD;
        x = x * x % MOD;
    }
    return res;
}

int edge[105][105];
int d[105][105];
int INF = 1000000;

void init(void){
    for (int i = 0; i < 105; i++){
        for (int j = 0; j < 105; j++){
            d[i][j] = INF;
            edge[i][j] = INF;
        }
    }
}

int main(void){
    init();
    int N, M;
    cin >> N >> M;
    for (int i = 0; i < M; i++){
        int a, b, c;
        cin >> a >> b >> c;
        edge[a][b] = c;
        edge[b][a] = c;
        d[a][b] = c;
        d[b][a] = c;
    }
    for (int k = 1; k <= N; k++){
        for (int i = 1; i <= N; i++){
            for (int j = 1; j <= N; j++){
                if (d[i][k] + d[k][j] < d[i][j]){
                    d[i][j] = d[i][k] + d[k][j];
                }
            }
        }
    }
    int ans = 0;
    for (int i = 1; i <= N; i++){
        for (int j = 1; j <= N; j++){
            if (edge[i][j] == INF) continue;
            if (edge[i][j] > d[i][j]) ans++;
        }
    }
    ans /= 2;
    cout << ans << endl;
    return 0;
}