zoukankan      html  css  js  c++  java
  • Sharing

    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 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

      1 //话说是12年408数据结构的真题唉~  输出格式 是个坑
      2 
      3  
      4 
      5 #include <iostream>
      6 
      7 #include <iomanip>
      8 
      9 #include <stdio.h>
     10 
     11 using namespace std;
     12 
     13  
     14 
     15 int cc[100001];
     16 
     17  
     18 
     19 int getlen(int x)
     20 
     21 {
     22 
     23     int count=0;
     24 
     25       while(x!=-1)
     26 
     27       {
     28 
     29          count++;
     30 
     31          x=cc[x];
     32 
     33       }
     34 
     35       return count;
     36 
     37 }
     38 
     39  
     40 
     41 int main()
     42 
     43 {
     44 
     45  
     46 
     47      int h1,h2,n;
     48 
     49       while(cin>>h1)
     50 
     51       {
     52 
     53          cin>>h2>>n;
     54 
     55  
     56 
     57          int tem=n;
     58 
     59          int add,next;
     60 
     61          char aa;
     62 
     63  
     64 
     65          while(tem--)
     66 
     67          {
     68 
     69             scanf("%d %c %d",&add,&aa,&next);
     70 
     71               cc[add]=next;
     72 
     73          }
     74 
     75  
     76 
     77        int l1,l2;
     78 
     79          l1=getlen(h1);
     80 
     81          l2=getlen(h2);
     82 
     83  
     84 
     85          while(l1>l2)
     86 
     87          {
     88 
     89             h1=cc[h1];
     90 
     91               l1--;
     92 
     93          }
     94 
     95  
     96 
     97           while(l1<l2)
     98 
     99          {
    100 
    101             h2=cc[h2];
    102 
    103               l2--;
    104 
    105          }
    106 
    107  
    108 
    109          while(h1!=h2)
    110 
    111          {
    112 
    113                h1=cc[h1];
    114 
    115                h2=cc[h2];
    116 
    117          }
    118 
    119  
    120 
    121            if(h1==-1) cout<<h1<<endl;
    122 
    123             else  cout<<setfill('0')<<setw(5)<<h1<<endl;
    124 
    125       }
    126 
    127   return 0;
    128 
    129 }
    130 
    131  
    View Code
  • 相关阅读:
    slurm.conf系统初始配置
    MySQL数据库服务器(YUM)安装
    Slurm任务调度系统部署和测试(源码)(1)
    并行管理工具——pdsh
    Munge服务部署和测试
    NTP服务部署和测试
    LDAP-openldap服务部署和测试(YUM安装)
    KVM虚拟机管理——虚拟机创建和操作系统安装
    KVM虚拟机管理——虚拟机克隆
    KVM虚拟化环境准备
  • 原文地址:https://www.cnblogs.com/xiaoyesoso/p/4249342.html
Copyright © 2011-2022 走看看