a1032 sharing

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).

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.

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.

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 

67890 

00001 00002 4 
00001 a 10001 
10001 s -1 
00002 a 10002 
10002 t -1 

-1

#include <bits/stdc++.h>
using namespace std;
const int max_n = 100001;

struct Node
{
    char data;
    int next;
    bool flag; 
} node[max_n];

int main()
{
    //初始化node中的flag
    for (int i = 0; i < max_n; i++)
        node[i].flag = false;
    int list_1, list_2, N;
    scanf("%d%d%d", &list_1, &list_2, &N);
    int addr, next;
    char data;
    for (int i = 0; i < N; i++)
    {
        scanf("%d %c %d", &addr, &data, &next);
        node[addr].data = data;
        node[addr].next = next;
    }

    int p = list_1; //遍历list_1
    for (; p != -1; p = node[p].next)
        node[p].flag = true; //遍历过的结点,置为true
    //遍历list_2
    for (p = list_2; p != -1; p = node[p].next)
        if (node[p].flag) //找到第一条链表中出现的结点
            break;
    if (p != -1) //没达到末尾,找到了公共结点
        printf("%05dn", p);
    else
        printf("-1n");
    return 0;
}