【思路】先求出两个链表的长度，假设长度分别为l1、l2(l1 > l2), 较长的从l1-l2+1开始遍历，最终它们一定会相遇。

#include<iostream>
using namespace std;
const int N = 100010;
int ne[N];
char cha[N];

int main()
{
    int b1, b2, n;
    cin >> b1 >> b2 >> n;
    for(int i = 0; i < n; i++)
    {
        int addre, next;
        char c;
        cin >> addre >> c >> next;
        ne[addre] = next;
        cha[addre] = c;
    }
    // 求两链表的长度
    int len1 = 0, len2 = 0;
    for(int i = b1; i != -1; i = ne[i]) len1++;
    for(int i = b2; i != -1; i = ne[i]) len2++;
    int p1 = b1, p2 = b2;
    if(len1 > len2)
    {
        while(len1 - len2 > 0)
        {
            p1 = ne[p1];
            len1--;
        }
    }
    else {
        while(len2 - len1 > 0)
        {
            p2 = ne[p2];
            len2--;
        }
    }

    while(p1 != -1 && p2 != -1)
    {
        if(p1 == p2)  break;
        p1 = ne[p1];
        p2 = ne[p2];
    }
    if(p1 != -1 && p2 != -1)    printf("%05d\n", p1);
    else puts("-1");
    return 0;
}