1032 Sharing (25 分)

最新推荐文章于 2025-04-26 22:24:20 发布

Tropine

最新推荐文章于 2025-04-26 22:24:20 发布

阅读量121

点赞数

分类专栏： PAT 甲级文章标签： c语言数据结构 c++ PAT

本文链接：https://blog.youkuaiyun.com/Tropine/article/details/120596700

版权

PAT 甲级专栏收录该内容

34 篇文章

订阅专栏

该博客讨论了一种使用链表存储英文单词的方法，通过共享子列表节省空间。当两个单词有相同的后缀时，它们可以共享存储。题目要求找到公共后缀的起始位置，如果不存在公共后缀则输出-1。博客提供了C++代码示例，通过遍历链表寻找共享节点来解决此问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1032 Sharing (25 分)

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.

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

whereAddress 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 Nextis 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

Analysis

这道题如果有链表的方法做的话会很麻烦，只能靠遍历来确定插入的位置，所以用数组来模拟链表。

Code(C++)

#include <iostream>
#include <iomanip>
using namespace std;

struct node
{
    char element;
    int next;
    bool flag;
};

int main()
{
    struct node a[100001];
    
    long long s1, s2;
    int n;
    cin >> s1 >> s2 >> n;
    
    for(int i=0; i<n; i++)
    {
        int p, q;
        char c;
        cin >> p >> c >> q;
        a[p].element = c;
        a[p].next = q;
        a[p].flag = false;
    }
    
    for(int i=s1; i!=-1; i=a[i].next)
    {
        a[i].flag = true;
    }
    for(int i=s2; i!=-1; i=a[i].next)
    {
        if(a[i].flag)
        {
            printf("%05d", i);
            return 0;
        }
    }
    cout << -1;
    
    return 0;
}