zoukankan      html  css  js  c++  java
  • PAT 1032. Sharing

    1032. Sharing (25)

    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:
    67890
    
    Sample Input 2:
    00001 00002 4
    00001 a 10001
    10001 s -1
    00002 a 10002
    10002 t -1
    
    Sample Output 2:
    -1

    分析
    这道题只要在word2找到出现过在word1中的字母即可,若没有,则输出-1。

     1 #include<iostream>
     2 using namespace std;
     3 struct Node{
     4     int flag=0,next;
     5 }; 
     6 Node list[100000];
     7 int main(){
     8     int s1,s2,n,id;
     9     char data;
    10     cin>>s1>>s2>>n;
    11     for(int i=0;i<n;i++){
    12         cin>>id>>data;
    13         cin>>list[id].next;
    14     }
    15     while(s1!=-1){
    16         list[s1].flag=1;
    17         s1=list[s1].next;
    18     }
    19     while(s2!=-1){
    20         if(list[s2].flag==1){
    21             printf("%05d",s2); 
    22             return 0;
    23         }
    24         s2=list[s2].next;
    25     }
    26     cout<<-1;
    27     return 0;
    28 } 
  • 相关阅读:
    用Struts2框架报错:The Struts dispatcher cannot be found
    Struts2.0笔记二
    [转]使用Struts 2防止表单重复提交
    Struts配置文件报错"元素类型为 "package" 的内容必须匹配"
    Java基础知识
    Struts笔记二:栈值的内存区域及标签和拦截器
    [转]迭代器
    Struts笔记一
    账户注册激活邮件及登入和注销
    EL表达式获取对象属性的原理
  • 原文地址:https://www.cnblogs.com/A-Little-Nut/p/8289604.html
Copyright © 2011-2022 走看看