用c++定义一有n个结点的的以邻接矩阵为存储方式的网（有向或无向均可），并对其进行如下操作：1.深度优先遍历，并输出遍历序列；2.广度优先遍历，并输出遍历序列；3.求源点到图中每个点的最短路径，并输出；

好的，以下是完整的C++代码实现：

#include <iostream>
#include <vector>
#include <queue>
#include <climits>
using namespace std;

struct Node {
    int id;
    int value;
};

class Graph {
public:
    Graph(int n);
    void addEdge(int u, int v, int w);
    void dfs();
    void bfs();
    void shortestPath(int s);
private:
    int n;
    vector<Node> nodes;
    vector<vector<int>> adjMatrix;
    void dfsHelper(int u, vector<bool>& visited);
};

Graph::Graph(int n) {
    this->n = n;
    for (int i = 0; i < n; i++) {
        nodes.push_back({i, 0});
    }
    adjMatrix.resize(n, vector<int>(n, 0));
}

void Graph::addEdge(int u, int v, int w) {
    adjMatrix[u][v] = w;
}

void Graph::dfs() {
    vector<bool> visited(n, false);
    for (int i = 0; i < n; i++) {
        if (!visited[i]) {
            dfsHelper(i, visited);
        }
    }
}

void Graph::dfsHelper(int u, vector<bool>& visited) {
    visited[u] = true;
    cout << nodes[u].id << " ";
    for (int v = 0; v < n; v++) {
        if (adjMatrix[u][v] != 0 && !visited[v]) {
            dfsHelper(v, visited);
        }
    }
}

void Graph::bfs() {
    vector<bool> visited(n, false);
    queue<int> q;
    for (int i = 0; i < n; i++) {
        if (!visited[i]) {
            q.push(i);
            visited[i] = true;
            while (!q.empty()) {
                int u = q.front();
                q.pop();
                cout << nodes[u].id << " ";
                for (int v = 0; v < n; v++) {
                    if (adjMatrix[u][v] != 0 && !visited[v]) {
                        q.push(v);
                        visited[v] = true;
                    }
                }
            }
        }
    }
}

void Graph::shortestPath(int s) {
    vector<int> dist(n, INT_MAX);
    dist[s] = 0;
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
    pq.push({0, s});
    while (!pq.empty()) {
        int u = pq.top().second;
        pq.pop();
        for (int v = 0; v < n; v++) {
            if (adjMatrix[u][v] != 0) {
                int newDist = dist[u] + adjMatrix[u][v];
                if (newDist < dist[v]) {
                    dist[v] = newDist;
                    pq.push({newDist, v});
                }
            }
        }
    }
    for (int i = 0; i < n; i++) {
        cout << "Shortest distance from " << s << " to " << nodes[i].id << " is " << dist[i] << endl;
    }
}

int main() {
    int n = 5;
    Graph g(n);
    g.addEdge(0, 1, 10);
    g.addEdge(0, 3, 5);
    g.addEdge(1, 2, 1);
    g.addEdge(1, 3, 2);
    g.addEdge(2, 4, 4);
    g.addEdge(3, 1, 3);
    g.addEdge(3, 2, 9);
    g.addEdge(3, 4, 2);
    g.addEdge(4, 0, 7);
    g.addEdge(4, 2, 6);
    g.dfs();
    cout << endl;
    g.bfs();
    cout << endl;
    g.shortestPath(0);
    return 0;
}

这个例子中，我们创建了一个有5个节点的图，并添加了10条边。然后分别进行了深度优先遍历、广度优先遍历和求最短路径的操作。输出结果如下：

0 1 2 4 3 
0 1 3 2 4 
Shortest distance from 0 to 0 is 0
Shortest distance from 0 to 1 is 8
Shortest distance from 0 to 2 is 9
Shortest distance from 0 to 3 is 5
Shortest distance from 0 to 4 is 7