【PAT】【Advanced Level】1032. Sharing (25)

最新推荐文章于 2022-05-05 19:26:29 发布

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最新推荐文章于 2022-05-05 19:26:29 发布

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该博客主要介绍了PAT甲级试题1032题目的解题思路和解决方案。题目要求通过链表结构存储英文单词，并找出两个单词的共享后缀起始位置。博主提供了实现方法，包括按要求构建链表，分别遍历并记录节点访问次数，然后再次遍历找出首次出现被访问两次的节点作为共享后缀的起始位置。

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1032. Sharing (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

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 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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1032

思路：

按要求组链表

分别对1,2进行遍历，同时记录每一个地址的遍历次数

任选一个进行二次遍历，找到第一个已经被遍历过两次的节点就是结果

CODE：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cstdlib>
using namespace std;

typedef struct S
{
    char val;
    int nxt;
};


S li1[100010];
int li[100010];

int main()
{
    memset(li,0,sizeof(li));
    string a,b;
    cin>>a>>b;
    int ad1=atoi(a.c_str());
    int ad2=atoi(b.c_str());
    int n;
    cin>>n;
    for (int i=0;i<n;i++)
    {
        char ad[50];
        char value;
        char nxt[50];
        scanf("%s %c %s",&ad,&value,&nxt);
        //dcin>>ad>>value>>nxt;
        li1[atoi(ad)].val=value;
        li1[atoi(ad)].nxt=atoi(nxt);
        //cout<<ad<<value<<nxt<<endl;
    }
    while (ad1 != -1)
    {
        //cout<<ad1<<endl;
        li[ad1]++;
        ad1=li1[ad1].nxt;
    }
    while (ad2 != -1)
    {
        //cout<<ad2<<endl;
        li[ad2]++;
        if (li[ad2]==2) break;
        ad2=li1[ad2].nxt;
    }
    int fl=0;
    for (int i=0;i<100000;i++)
    {
        if (li[i]==2)
        {
            fl=1;
            printf("%05d\n",i);
            //cout<<i<<endl;
            break;
        }
    }
    if (fl==0)
        cout<<-1;
    return 0;
}