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

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