1032 Sharing (25 分)
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
- whereAddress 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
大意
求两条链表的首个共同节点的地址, 没有输出 - 1.
输入:
? 第一行 包含 两个地址 正整数 N
? (两个地址) 两个单词首个节点的地址
? (正数 N) 节点总数量
? 后面 N 行 Address Data Next
节点地址是 5 位正整数,-1 代表 NULL
[例子 1]
单词 1:LoaDing
单词 2:Being
输出 i 的地址
分析
用结构体存储节点, data 是英文字母, next 是下一个节点的地址, flag 记录第一条链表是否访问过该节点.
遍历第一条链表, 将访问过的节点的 flag 赋值为 true. 遍历第二条链表, 当碰到 flag 为 true 的节点即找到了.
注意
虽然题目是链表处理, 但实现程序不一定会用到链表.
输出必须为 5 位数, 需要规定格式
printf("%05d\n",i);
这里不能写成以下 (部分 AC)
printf("%d\n",i);
代码
- #include<iostream>
- using namespace std;
- // 节点结构体
- struct NODE {
- char data;
- int next;
- bool flag;// 第一条链表若访问过为 true
- } n[100000];
- int main() {
- int s1,s2,num;// 两个单词的起始地址, 节点总个数
- scanf("%d %d %d",&s1,&s2,&num);
- /* 赋初值 */
- int addr,next;
- char data;
- for(int i=0; i<num; i++) {
- scanf("%d %c %d",&addr,&data,&next);
- n[addr]= {data,next,false};
- }
- /* 第一条链表 */
- for(int i=s1; i!=-1; i=n[i].next) {
- n[i].flag=true;
- }
- /* 第二条链表 */
- for(int i=s2; i!=-1; i=n[i].next) {
- if(n[i].flag==true) {
- printf("%05d\n",i);
- return 0;
- }
- }
- printf("-1");
- return 0;
- }
单词
suffix: 后缀
来源: http://www.bubuko.com/infodetail-3394785.html