Question

我正在尝试用C ++实现图形和广度优先遍历算法。这是我关注的代码

#include<iostream>
#include <list>

using namespace std;
class Graph
{
int V;    // No. of vertices
list<int> *adj;    // Pointer to an array containing adjacency lists
public:
Graph(int V);  // Constructor
void addEdge(int v, int w); // function to add an edge to graph
bool isReachable(int s, int d);  // returns true if there is a path from s to d
};

Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}

void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}

// A BFS based function to check whether d is reachable from s.
bool Graph::isReachable(int s, int d)
{
// Base case
if (s == d)
  return true;

// Mark all the vertices as not visited
bool *visited = new bool[V];
for (int i = 0; i < V; i++)
    visited[i] = false;

// Create a queue for BFS
list<int> queue;

// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);

// it will be used to get all adjacent vertices of a vertex
list<int>::iterator i;

while (!queue.empty())
{
    // Dequeue a vertex from queue and print it
    s = queue.front();
    queue.pop_front();

    // Get all adjacent vertices of the dequeued vertex s
    // If a adjacent has not been visited, then mark it visited
    // and enqueue it
    for (i = adj[s].begin(); i != adj[s].end(); ++i)
    {
        // If this adjacent node is the destination node, then return true
        if (*i == d)
            return true;

        // Else, continue to do BFS
        if (!visited[*i])
        {
            visited[*i] = true;
            queue.push_back(*i);
        }
    }
}

return false;
}

// Driver program to test methods of graph class
int main()
{
// Create a graph given in the above diagram

Graph g(23);
g.addEdge(1, 2);
g.addEdge(2, 3);
g.addEdge(3, 4);
g.addEdge(4, 5);
g.addEdge(5, 6);
g.addEdge(6, 7);
g.addEdge(7, 8);
g.addEdge(8, 9);
g.addEdge(9, 10);
g.addEdge(10, 11);
g.addEdge(11, 12);
g.addEdge(12, 13);
g.addEdge(13, 14);
g.addEdge(14, 15);
g.addEdge(15, 16);
g.addEdge(8, 17);
g.addEdge(17, 18);
g.addEdge(18, 19);
g.addEdge(19, 20);
g.addEdge(20, 21);
g.addEdge(21, 22);
g.addEdge(22, 23);
g.addEdge(23, 16);


int u = 1, v = 3;
if(g.isReachable(u, v))
    cout<< "\n There is a path from " << u << " to " << v;
else
    cout<< "\n There is no path from " << u << " to " << v;

u = 3, v = 1;
if(g.isReachable(u, v))
    cout<< "\n There is a path from " << u << " to " << v;
else
    cout<< "\n There is no path from " << u << " to " << v;

return 0;
}

在主方法中，如果我包括添加到26，则假设它在＃4; 0x4A43BC7中的未处理异常在Implementation4.exe中：0xC0000005：访问冲突读取位置0xABABABAF。＆＃34;异常。

无法弄清楚我在这里缺少什么。任何人都可以帮我解决这个问题。

Answer 1

C ++中的索引是基于零的，而不是基于索引的。当您呼叫addGraph时，您使用索引1到23调用它。最后一次调用addGraph会导致它尝试将值添加到第24个 list在只有 23 列表的数组中。要修改此更改，调用addGraph将从0开始并在22结束，或者使用24个列表的数组对其进行初始化。

另外，我建议您使用裸列表列表切换到使用std::vector<list>。这将解决内存泄漏，因为您没有删除adj的析构函数。

未处理的异常：实现图形时访问冲突读取位置错误

1 个答案: