1032 Sharing (25分)

1 题目

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 (≤10^​5​​ ), 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 -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:
67890

Sample Input 2:
00001 00002 4
00001 a 10001
10001 s -1
00002 a 10002
10002 t -1
  
Sample Output 2:
-1

2 解析

• 题意： 找出两条链表第一个公共的结点地址输出，如果没有则输出-1
• 分析

• 使用静态链表（地址范围小）存储

• 结点的数据、下一个结点地址、是否在第一条链表出现过

• 从头遍历第一条链表，把该链表的每个结点中的flag标记为true
• 从头遍历第二条链表，如果该结点的flag为true，记录下此结点的地址，跳出循环
• 输出第一个公共结点的地址

• 坑：

• 输出位数，不满5位时，用%05d
• 读入字母时候，%c可以读入空格,输入地址、数据、下一个结点地址的时候，不能写成%d%c%d,应该在他们中间加空格
• 使用map会造成超时

• 出现的错误：

• ​​scanf("%d %c %d", &adress, &node[adress].data, &node[adress].next);​​ 输入的数据，没有输入完，则不能拿来用

3 参考代码

#include 

#define MAXN 100010

struct NODE
{
  char data;//数据
  int next;//下一个地址
  bool flag;//是否在第一条链表中出现国
}node[MAXN];//静态链表

int main(int argc, char const *argv[])
{
  for (int i = 0; i != MAXN; ++i)//初始化flag
  {
    node[i].flag = false;
  }

  int s1, s2, n;//记录输入的头结点，以及待输入的行数
  scanf("%d%d%d", &s1, &s2, &n);

  int adress,next;//结点、后继结点
  char data;
  while(n--){
    scanf("%d %c %d", &adress, &data, &next);
    node[adress].data = data;
    node[adress].next = next;
  }

  int ad;//当前查找的地址
  for (ad = s1; ad != -1; ad = node[ad].next)
  {
    node[ad].flag = true;//如果在第一个链表中出现过，则flag标记为true
  }

  for (ad = s2; ad != -1; ad = node[ad].next)
  {
      //在第二个链表中查找出现的第一个链表中的结点
    if(node[ad].flag == true){
      break;
    }
  }

  if(ad == -1){//如果查找结点的地址为空结点，则没有公共结点
    printf("-1\n");
  }else{
    printf("%05d\n", ad);
  }

 return 0;

}