标签：
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
标签：
原文地址：http://www.cnblogs.com/xiaoyesoso/p/4249342.html
