Question

我有一个无向图，需要用邻接表来表示它。我正在使用单个链接列表的数组。输入为：

11 18 3
1 8 4 7 7 10 11 10 2 1 2 3 8 9 8 3 9 3 9 2 5 6 5 11 1 4 10 6 7 6 2 8 11 7 11 6

首先，我正在创建列表数组。然后，要将节点添加到其中，我从输入中读取并创建2个指针。第一个保留必须添加到列表中的值作为信息，而第二个保留为列表的当前最后一个元素。然后，我将第一个绑定到第二个，第一个绑定到列表的最后一个元素。由于图形是无向的，因此我现在删除指针并创建新的指针以添加到反向对中。

输出不一致，并且从不匹配预期的输出。

#include <iostream>
#include <fstream>
using namespace std;
ifstream f("info.in");
struct LLSI {
    int info;
    LLSI* urm;
} * c, *d, *cap, *v[100], *e;
int main()
{
    int n, m, k, i, x, y;
    f >> n >> m >> k;
  //creating the array of lists
  for (i = 1; i <= n; i++) {
        v[i] = new LLSI;
        v[i]->urm = 0;
        v[i]->info = 0;
    }
    //adding the nodes into the adjacency list
    for (i = 1; i <= m; i++) {
        f >> x >> y;
        c = new LLSI;
        c->info = y;
        c->urm = 0;
        d = new LLSI;
        d = v[x];
        while (d->urm)
            d = d->urm;
        d->urm = c;
        v[x]->info++;
        d = c;
        delete c;
        delete d;
        d = new LLSI;
        d = v[y];
        while (d->urm)
            d = d->urm;
        c = new LLSI;
        c->info = x;
        c->urm = 0;
        d->urm = c;
        v[y]->info++;
        delete c;
        delete d;
    }
    //outputting the adjancecy list
    for (i = 1; i <= n; i++) {
        c = new LLSI;
        c = v[i];
        while (c) {
            cout << c->info << " ";
            c = c->urm;
        }
        cout << "\n";
        delete c;
    }
    return 0;
}

Answer 1

此

//outputting the adjancecy list
for (i = 1; i <= n; i++) {
    c = new LLSI;
    c = v[i];
    while (c) {
        cout << c->info << " ";
        c = c->urm;
    }
    cout << "\n";
    delete c;
}

应该是这个

//outputting the adjancecy list
for (i = 1; i <= n; i++) {
    c = v[i];
    while (c) {
        cout << c->info << " ";
        c = c->urm;
    }
    cout << "\n";
}

我删除了new和delete。不知道为什么您认为需要创建一个新列表以遍历列表。

在先前的循环中，您有相同的错误。这个

//adding the nodes into the adjacency list
for (i = 1; i <= m; i++) {
    f >> x >> y;
    c = new LLSI;
    c->info = y;
    c->urm = 0;
    d = new LLSI;
    d = v[x];
    while (d->urm)
        d = d->urm;
    d->urm = c;
    v[x]->info++;
    d = c;
    delete c;
    delete d;
    d = new LLSI;
    d = v[y];
    while (d->urm)
        d = d->urm;
    c = new LLSI;
    c->info = x;
    c->urm = 0;
    d->urm = c;
    v[y]->info++;
    delete c;
    delete d;
}

应该是这个

//adding the nodes into the adjacency list
for (i = 1; i <= m; i++) {
    f >> x >> y;
    c = new LLSI;
    c->info = y;
    c->urm = 0;
    d = v[x];
    while (d->urm)
        d = d->urm;
    d->urm = c;
    v[x]->info++;
    delete c;
    d = v[y];
    while (d->urm)
        d = d->urm;
    c = new LLSI;
    c->info = x;
    c->urm = 0;
    d->urm = c;
    v[y]->info++;
    delete c;
}

我删除了d = new LLSI;和delete d;。使用new创建一个新对象。如果只想指向一个现有对象，则不需要new或delete。

另一个明显的改进是您有很多重复的代码，它们只是在列表x或列表y上进行操作而有所不同。重复的代码应放在单独的函数中。

Answer 2

谢谢您的建议，出于某种原因，我的老师教我我应该使用new和delete，尽管这样做更有意义。该代码现在未显示任何输出，但这是新代码的样子：

#include <iostream>
#include <fstream>
using namespace std;
ifstream f("info.in");
struct LLSI {
    int info;
    LLSI* urm;
} * c, *d, *cap, *v[100], *e;
int main()
{
    int n, m, k, i, x, y;
    f >> n >> m >> k;
    for (i = 1; i <= n; i++) {
        v[i] = new LLSI;
        v[i]->urm = 0;
        v[i]->info = 0;
    }
    for (i = 1; i <= m; i++) {
        f >> x >> y;
        c = new LLSI;
        c->info = y;
        c->urm = 0;
        d = v[x];
        while (d->urm)
            d = d->urm;
        d->urm = c;
        v[x]->info++;
        delete c;
        d = v[y];
        while (d->urm)
            d = d->urm;
        c = new LLSI;
        c->info = x;
        c->urm = 0;
        d->urm = c;
        v[y]->info++;
        delete c;
    }
    for (i = 1; i <= n; i++) {
        c = v[i];
        while (c) {
            cout << c->info << " ";
            c = c->urm;
        }
        cout << "\n";
    }
    return 0;
}

创建邻接表时输出不一致

2 个答案: