1032. Sharing

最新推荐文章于 2021-06-08 10:46:42 发布

原创最新推荐文章于 2021-06-08 10:46:42 发布 · 159 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT

PAT (Advanced Level) 专栏收录该内容

45 篇文章

订阅专栏

本文介绍了一种算法，用于找到两个链表的第一个共用结点的位置。通过使用静态链表存储英文单词，并标记已访问结点的方式，有效地解决了这一问题。

Sharing

To store English words, one method is to use linked lists and store a word letter by letter. To save some space, we may let the words share the same sublist if they share the same suffix. For example, “loading” and “being” are stored as showed in Figure 1.

You are supposed to find the starting position of the common suffix (e.g. the position of “i” in Figure 1).

Input Specification:

Each input file contains one test case. For each case, the first line contains two addresses of nodes and a positive N (<= 105), where the two addresses are the addresses of the first nodes of the two words, and N is the total number of nodes. The address of a node is a 5-digit positive integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is the letter contained by this node which is an English letter chosen from {a-z, A-Z}, and Next is the position of the next node.

Output Specification:

For each case, simply output the 5-digit starting position of the common suffix. If the two words have no common suffix, output “-1” instead.

Sample Input 1:

11111 22222 9
67890 i 00002
00010 a 12345
00003 g -1
12345 D 67890
00002 n 00003
22222 B 23456
11111 L 00001
23456 e 67890
00001 o 00010

Sample Output 1:

Sample Input 2:

00001 00002 4
00001 a 10001
10001 s -1
00002 a 10002
10002 t -1

Sample Output 2:

-1

题意

给出两个链表的首地址及结点总数，并给出每个结点的地址、数据、下个结点地址，寻找两个链表第一个共用结点的地址，如果不存在则输出-1。

思路

地址的长度固定且数据范围不大，应当用静态链表处理。题目实际上与结点内数据没有关系，只需要考虑结点本身，因此可以采取下列措施：在定义结构体时多用一个bool成员来判断该结点是否被访问过，首先遍历第一个链表，将每个结点标记为已访问，再遍历第二个链表，如果当前结点已被访问，则返回该结点地址；如果所有结点都没被访问过，则返回-1。

代码实现

#include <cstdio>

struct Node
{
    char ch;
    int next;
    bool flag;      // 判断是否被访问过
} node[100000];

int findPos(int a, int b)
{
    while (a != -1)     // 遍历第一个链表并标记已访问
    {
        node[a].flag = true;
        a = node[a].next;
    }

    while (b != -1)     // 查找第二个链表中是否有已访问结点
    {
        if (node[b].flag)
            return b;
        b = node[b].next;
    }

    return -1;
}

int main()
{
    int a, b;
    int n;
    int x, y;
    char c;
    int pos;

    scanf("%d %d %d", &a, &b, &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d %c %d", &x, &c, &y);
        node[x].ch = c;
        node[x].next = y;
    }

    pos = findPos(a, b);
    if (pos != -1)
        printf("%05d", pos);         // 注意格式输出
    else
        printf("%d", pos);

    return 0;
}