1032. Sharing (25)

最新推荐文章于 2023-08-04 09:37:38 发布

原创最新推荐文章于 2023-08-04 09:37:38 发布 · 970 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT #链表 #公共节点

PAT 专栏收录该内容

104 篇文章

订阅专栏

本文详细解析PAT-A Practise中的Sharing(25)题目，介绍如何通过链表存储英文单词并利用公共前缀节省空间。通过输入输出样例分析，提供一种寻找公共前缀起点的算法实现方式，同时讨论代码长度限制与判题程序的影响。

题目链接：http://www.patest.cn/contests/pat-a-practise/1032

题目：

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}, andNext 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

分析：

两个链表，找到它们相交的第一个节点。

注意：

可能存在多个字串，并且可能有其他非题目中两字串的节点存在，并且注意最后是要输出%05d的

AC代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int nodes[100001];
int real_nodes[100001];
int main(void){
 //freopen("F://Temp/input.txt", "r", stdin);
 int s1, s2, n;
 scanf("%d %d %d", &s1, &s2, &n);
 int s, t;
 char c;
 for (int i = 0; i < 100001; i++){
  nodes[i] = 0;
  real_nodes[i] = 0;
 }
 for (int i = 0; i < n; i++){
  scanf("%d %c %d", &s, &c, &t);
  nodes[s] = t;
 }
 int start = s1; int end;
 while (start != -1){
  real_nodes[start] = 1;
  start = nodes[start];
 }//找到第一个链表的所有节点，并置为1
 start = s2;
 bool find = false;
 while (start != -1){
  if (real_nodes[start] != 0){//如果第二个链表中地址已经有记录，则就是第一个公共节点
   printf("%.5d\n", start);
   return 0;
  }
  else {
   real_nodes[start] = nodes[start];
   start = nodes[start];
  }
 }
 puts("-1");
 return 0;
}

截图：

——Apie陈小旭