【发布时间】:2015-02-22 21:56:15
【问题描述】:
我在这里使用邻接表来表示一个图。
代码如下:
// A program to check the reachability between two nodes within a specified number of steps using an adjacency list
/***************************************************************************/
#include <iostream>
#include <fstream>
using namespace std;
struct node {
int value;
node* next;
// Constructors:
node() {
value = 0;
next = 0;
}
node(int x) {
value = x;
next = 0;
}
};
node* al; // Adjacency list
int n; // Number of nodes
int node1, node2, k; // For reading input from the console
int counter;
bool checkReachability(int, int, int);
void freeMemory();
int main() {
ifstream in;
in.open("Input.txt");
if(in) {
in >> n;
al = new node[n];
for (int i = 0; i < n; i++) {
al[i].value = i+1;
}
int a, b;
while(in >> a >> b) {
node* temp = &al[a-1];
while(temp->next != 0) {
temp = temp->next;
}
temp->next = new node(b);
}
cout << "\n\nThe adjacency list representation of the graph is as follows: \n";
cout << "________________________________\n\n";
for (int i = 0; i < n; i++) {
cout << al[i].value;
node* temp = al[i].next;
while(temp != 0) {
cout << "->" << temp->value;
temp = temp->next;
}
cout << endl;
}
cout << "________________________________\n";
in.close();
char c;
do {
cout << "\nPlease enter the input (node1, node2, k): \n";
cin >> node1 >> node2 >> k;
counter = 0;
if (checkReachability(node1 - 1, node2, k)) {
cout << "\nReachable within " << k << " steps";
if (counter < k) {
cout << " (actually " << counter << ")";
}
cout << endl << endl;
}
else {
cout << "\nNot reachable within " << k << " steps \n";
}
cout << "\nDo you want to continue? Y/N \n\n";
cin >> c;
} while (c == 'Y' || c == 'y');
freeMemory();
} else {
cout << "\nCouldn't find the input file\n\n";
}
return 0;
}
bool checkReachability(int n1, int n2, int k) {
if ((n1 + 1) == n2) return true;
counter++;
if (counter <= k) {
node* temp = &(al[n1]);
while (temp != 0) {
if (temp->value == n2) return true;
temp = temp->next;
}
temp = al[n1].next;
while (temp != 0) {
if (checkReachability(((temp->value)-1),n2,k)) return true;
counter--;
temp = temp->next;
}
}
return false;
}
void freeMemory() {
cout << "\nFreeing memory...\n";
// To free the dynamically allocated memory on the heap
for (int i = 0; i < n; i++) {
node* temp = &al[i];
while(temp != 0) {
node* temp2 = temp;
temp = temp->next;
delete temp2;
}
}
//delete [] al;
cout << "\nMemory freed.\n";
}
程序运行良好。只有当我选择退出它时,它才会调用它崩溃的 freeMemory 函数。请帮我找出问题所在。
Input.txt 文件:
5
1 2
2 5
3 4
1 3
输出:
The adjacency list represent
____________________________
1->2->3
2->5
3->4
4
5
____________________________
Please enter the input (node
1 2 1
Reachable within 1 steps
Do you want to continue? Y/N
y
Please enter the input (node
2 4 4
Not reachable within 4 steps
Do you want to continue? Y/N
N
Freeing memory...
然后,它崩溃了。
【问题讨论】:
-
双重
delete我怀疑。另请参阅:stackoverflow.com/questions/28649952/… -
也可以粘贴输入文件内容,最好是可以重现您的问题的合理内容。请添加到问题中。包含崩溃的输出也不会不受欢迎。
-
@WhozCraig:好的,刚刚添加了 Input.txt 文件。
-
我怀疑您的清理循环是
delete数组中的初始条目,它是 not 动态分配的。只有通过next链接 的条目与new独立分配。a[]的初始持有是向量分配。我会在那里检查。 -
也添加了输出。
标签: c++