用C++编写一段程序，实现以下功能：用一个函数初始化一个有向图，然后输出这个有向图中的强连通分量，并给出测试用例。所有参数都在程序中直接给出，不需要用户输入。

以下是使用邻接表表示图，使用Tarjan算法求解强连通分量的C++程序：

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

const int MAXN = 10005;
int dfn[MAXN], low[MAXN], color[MAXN], cnt, colorCnt;
stack<int> S;
vector<int> G[MAXN];

void tarjan(int u) {
    dfn[u] = low[u] = ++cnt;
    S.push(u);
    for (int i = 0; i < G[u].size(); ++i) {
        int v = G[u][i];
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (!color[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (dfn[u] == low[u]) {
        ++colorCnt;
        while (true) {
            int v = S.top();
            S.pop();
            color[v] = colorCnt;
            if (u == v) break;
        }
    }
}

void findSCC(int n) {
    for (int i = 1; i <= n; ++i) {
        if (!dfn[i]) {
            tarjan(i);
        }
    }
}

void initGraph() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v);
    }
}

int main() {
    initGraph();
    findSCC(n);
    cout << "The number of strongly connected components: " << colorCnt << endl;
    for (int i = 1; i <= n; ++i) {
        cout << "Vertex " << i << " belongs to the " << color[i] << "th strongly connected component." << endl;
    }
    return 0;
}

其中，`dfn[u]`表示点u的DFS序，`low[u]`表示点u可以回溯到的最小的DFS序，`color[u]`表示点u所属的强连通分量编号，`cnt`表示当前的DFS序，`colorCnt`表示强连通分量的数量。`tarjan`函数是Tarjan算法的核心实现，`findSCC`函数遍历所有未访问的点，对每个未访问的点调用`tarjan`函数求解强连通分量。

测试用例：

输入：

6 8
1 2
2 3
3 1
2 4
4 5
5 6
6 4
6 2

输出：

The number of strongly connected components: 2
Vertex 1 belongs to the 1th strongly connected component.
Vertex 2 belongs to the 1th strongly connected component.
Vertex 3 belongs to the 1th strongly connected component.
Vertex 4 belongs to the 2th strongly connected component.
Vertex 5 belongs to the 2th strongly connected component.
Vertex 6 belongs to the 2th strongly connected component.