PAT-A 1032 Sharing （25 分)

最新推荐文章于 2022-08-12 15:43:56 发布

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最新推荐文章于 2022-08-12 15:43:56 发布

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本文介绍了一种通过链表存储英文单词并查找共同后缀的算法。利用节点地址映射和栈操作，算法能高效地找到两个单词的最长公共后缀起始位置，适用于空间节省的场景。

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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 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 -1instead.

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

思路：数据部分没有用到，因此可以将每个节点保存到map中，映射是从该节点地址到下一节点的地址。将两个单词的结点的地址依次入栈，再依次弹出，直到有一个栈为空或者弹出的值不等为止。输出的时候注意，如果不是-1要使用"%05d"进行格式化。

#include <cstdio>
#include <map>
#include <stack>

using namespace std;

int main()
{
    int head1, head2, n;
    scanf("%d %d %d", &head1, &head2, &n);

    map<int, int> all;  // address -> next

    for(int i = 0; i < n; i++)
    {
        int address, next;
        char data;

        scanf("%d %c %d", &address, &data, &next);
        all.insert(make_pair(address, next));
    }

    stack<int> word1, word2;

    while(head1 != -1)
    {
        word1.push(head1);
        head1 = all[head1];
    }

    while(head2 != -1)
    {
        word2.push(head2);
        head2 = all[head2];
    }

    int result = -1;
    while(!word1.empty() && !word2.empty() && word1.top() == word2.top())
    {
        result = word1.top();
        word1.pop();
        word2.pop();
    }
    if(result == -1)
        printf("-1\n");
    else
        printf("%05d\n", result);
    return 0;
}