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
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:
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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:
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1 |
67890 |
Sample Input 2:
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00001 00002 4 00001 a 10001 10001 s -1 00002 a 10002 10002 t -1 |
Sample Output 2:
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1 |
-1 |
分别遍历寻找公共节点即可。
代码如下：
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#include<stdio.h> typedef struct list_ { int word; //int key; int next; } list; list data[100000]; int main() { int n,i,address,head_a,head_b; char buf[2]; scanf("%d%d%d",&head_a,&head_b,&n); for(i=0;i<n;i++) { scanf("%d",&address); scanf("%s%d",buf,&(data[address].next)); data[address].word=0; } address=head_a; while(address!=-1) { data[address].word=1; address=data[address].next; } address=head_b; while(address!=-1) { if(data[address].word==1) { break; } address=data[address].next; } if(address==-1) { printf("-1\n"); } else { printf("%05d\n",address); } } |