PAT--1032 Sharing (25 分)

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

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:

Sample Input 2:

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

Sample Output 2:

-1

#include <bits/stdc++.h>
#include <iostream>

using namespace std;

const int maxn = 100010;

struct LIAN
{
    char data;
    int next;
    bool flag;  //结点是否在第一条链中出现过
}node[maxn];


int main()
{

    for(int i=0; i<=maxn; i++)
    {
        node[i].flag=false;
    }

    int s1,s2,n;
    cin>>s1>>s2>>n;
    int add,next;
    char data;
    for(int i=0; i<n; i++)
    {
        cin>>add>>data>>next;
        node[add].data = data;
        node[add].next = next;
    }

    int p;
    //将第一条链的flag设为true
    for(p=s1; p!=-1; p=node[p].next)
    {
        node[p].flag = true;
    }

    for(p=s2; p!=-1; p=node[p].next)
    {
        if(node[p].flag == true)
            break;
    }

    if(p!=-1)
    {
        printf("%05d\n",p);
    }
    else
    {
        cout<<"-1"<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}

思路（来自算法笔记）：

由题目给出的第一条链表的首地址出发遍历第一条链表，将经过的所有节点flag值设为TRUE；
接下来枚举第二条链表，当出现第一个flag值为true的结点，说明是第一条链表中出现过的结点；如果第二条遍历结束仍没有发现共用结点，则输出1；