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

• 由于地址的范围较小，可以使用静态链表。并根据题意，为链表节点结构体增加一个 flag 变量，用于表示该节点是否在第一条链表中，默认为 false。先由第一条链表的首地址出发，遍历第一条链表，并将其节点的 flag 都设为 true。当遍历第二条链表时，只需输出第一个 flag 为true 的节点的地址即可。

注意点

• 使用 %05d 格式输出地址，使其高位补 0。

#include <cstdio>

const int maxn = 100000;

struct Node {
    char data;
    int next;
    bool flag = false;
} nodes[maxn];

int main() {
    int root1, root2, n, address, next;
    char data;
    scanf("%d %d %d", &root1, &root2, &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &address);
        scanf(" %c %d", &nodes[address].data, &nodes[address].next);
    }
    int cur = root1;
    while (cur != -1) {
        nodes[cur].flag = true;
        cur = nodes[cur].next;
    }
    cur = root2;
    while (cur != -1) {
        if (nodes[cur].flag == true) {
            printf("%05d", cur);
            return 0;
        }
        cur = nodes[cur].next;
    }
    printf("-1");
    return 0;
}