PAT

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 Nextis 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 967890 i 0000200010 a 1234500003 g -112345 D 6789000002 n 0000322222 B 2345611111 L 0000123456 e 6789000001 o 00010

Sample Output 1:

67890

Sample Input 2:

00001 00002 400001 a 1000110001 s -100002 a 1000210002 t -1

Sample Output 2:

-1

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给定条件：
1.n个节点的地址，值，下一个地址
2.两个链表的第一个节点地址

要求：
1.这两个链表的最长公共后缀
2.输出这个后缀的第一个节点的地址

求解：
1.如果两个链遍又公共的后缀，那么如果遍历这两个链表，则这个后缀被访问过两次
2.遍历其中一个链表，并且标记访问过
3.遍历另一个链表，如果当前节点被遍历过，则这个节点（第一个被发现的）就是公共后缀的第一个节点

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#include <cstdio>using namespace std;struct Node{char val;int next;bool vis = false;} node[100010];int start1, start2, n, ad, next;char val;int main() {scanf("%d%d%d", &start1, &start2, &n);for(int i = 0; i < n; i++) {scanf("%d %c %d",&ad, &val, &next);node[ad].val = val;node[ad].next = next;}for(int i = start1; i != -1; i = node[i].next) {node[i].vis = true;}for(int i = start2; i != -1; i = node[i].next) {if(node[i].vis == true) {printf("%05d\n", i);return 0;}}printf("-1\n");return 0;}

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