算法笔记 P261 例题：【PAT A1032】Sharing

算法笔记练习 题解合集

本题链接

题目

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.

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:

-1

思路

更好的思路应该是建一个哈希表，在遍历第一条链表的时候将出现过的地址记录下来，然后遍历第二条链表，当遇到了第一次遍历过的地址即为所求。

---

而我被题意带偏了，先根据两条链表生成两个字符串，然后求两个字符串的最大公共后缀，再顺着链表找第一个公共结点…略微繁琐了一点。

注意如果用我的方法，求出字符串的最大公共后缀是不够的，例如对于loading和being，虽然最大公共后缀是i，但是有可能在两条链表里面i存在不同地址里，如果不注意这个细节，会卡在测试点5。

代码

#include 
#include 
#include 
#include 
using namespace std;
struct Node {
	char data;
	int next;
};
vector<Node> nodes(100010);
string getWord(int head) {
	string ret;
	while (head != -1) {
		ret += nodes[head].data;
		head = nodes[head].next; 
	}
	return ret;
} 
int main() {
	int headA, headB, n, address, tail;
	string a, b;
	cin >> headA >> headB >> n;
	while (n--) {
		cin >> address;
		cin >> nodes[address].data >> nodes[address].next;
	} 
	a = getWord(headA);
	b = getWord(headB);
	for (tail = 0; tail < min(a.size(), b.size()); ++tail)
		if (a[a.size() - tail - 1] != b[b.size() - tail - 1])
			break;
	if (tail == 0) {
		puts("-1"); 
		return 0;
	}
	for (int i = 0; i < a.size() - tail; ++i)
		headA = nodes[headA].next;
	for (int i = 0; i < b.size() - tail; ++i)
		headB = nodes[headB].next;
	while (headA != -1 && headB != -1 && headA != headB) {
		headA = nodes[headA].next;
		headB = nodes[headB].next;
	}
	if (headA == -1)
		puts("-1");
	else 
		printf("%05d\n", headA); 
	return 0;
}