请用C++为我写一段代码，要求能实现以下功能：求出带权值的有向图里任意两点间的五条相对长度最短的路径

好的，以下是一份可行的C++代码，使用了Dijkstra算法和堆优化，时间复杂度为O(mlogn)：

#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 1005;
const int MAXM = 100005;

struct Edge {
    int from, to, w;
    Edge(int from, int to, int w) : from(from), to(to), w(w) {}
};

struct Node {
    int id, dis;
    bool operator<(const Node& rhs) const {
        return dis > rhs.dis;
    }
};

vector<Edge> edges;
vector<int> G[MAXN];
int n, m;
int d[MAXN][6]; // 任意两点间相对长度最短的五条路径
bool vis[MAXN];

void dijkstra(int s) {
    memset(d, INF, sizeof(d));
    memset(vis, 0, sizeof(vis));
    priority_queue<Node> pq;
    pq.push({s, 0});
    d[s][0] = 0;
    while (!pq.empty()) {
        Node u = pq.top();
        pq.pop();
        if (vis[u.id]) continue;
        vis[u.id] = true;
        for (int i = 0; i < G[u.id].size(); i++) {
            Edge& e = edges[G[u.id][i]];
            for (int j = 0; j < 5; j++) {
                int v = e.to, w = e.w;
                if (d[u.id][j] == INF) break; // 之前的路径不存在，后面的也不存在
                if (d[v][j+1] > d[u.id][j] + w) {
                    d[v][j+1] = d[u.id][j] + w;
                    pq.push({v, d[v][j+1]});
                }
            }
        }
    }
}

int main() {
    // 输入n, m和边的信息
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        edges.push_back(Edge(u, v, w));
        G[u].push_back(i);
    }

    // 求任意两点间相对长度最短的五条路径
    for (int i = 1; i <= n; i++) {
        dijkstra(i);
        for (int j = 1; j <= n; j++) {
            if (i == j) continue;
            sort(d[j], d[j]+6);
            cout << "从节点 " << i << " 到节点 " << j << " 的五条最短路径长度分别为: ";
            for (int k = 1; k <= 5; k++) {
                if (d[j][k] == INF) break;
                cout << d[j][k] << " ";
            }
            cout << endl;
        }
    }

    return 0;
}

希望这份代码能够帮助到你！