1032 Sharing (25 point(s))

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简单题

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链表

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本文介绍了一种使用链表存储英文单词的方法，通过共享相同后缀的子链表来节省空间。主要讨论如何找到两个单词共有后缀的开始位置，通过比较节点地址并找出重复指向的地址实现。

问题描述：

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.
在这里插入图片描述
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 (≤10^5), 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:

首先比较是否是同一个单词，然后寻找是否有某一个地址被指向了两次即可。

#include<bits/stdc++.h>
using namespace std;
int lett1, lett2, n;
int main(){
	cin >> lett1 >> lett2 >> n;
	if (lett1 == lett2){
		printf("%05d", lett1);
		return 0;
	}
	int next[n];
	for(int i = 0; i < n; i ++){
		int ads, nxt;
		char key;
		scanf("%d %c %d", &ads, &key, &nxt);
		next[i] = nxt;
	}
	int suffix = -1;
	sort(next, next + n);
	for (int i = 0; i < n - 1; i ++){
		if (next[i] != -1 && next[i] == next[i + 1]){
			suffix = next[i];
		}
	}
	if(suffix == -1) cout << suffix;
	else printf("%05d", suffix);
	return 0;
}