题目

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:

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

刚看了静态链表这一节，挺好，让我写的话估计还想着用普通链表记录地址，这样再去扫描比较

这里很重要的思路就是一个节点只会有一个next，即使有多个pre。

也就是说，只要确定两条链表同时出现的第一个节点，它后面的内容就一定一样

结构体数组没法直接初始化（要初始化得遍历数组），不像类那样直接来个默认值，所以一次遍历链表是不可能完成的。

如果用类的话带个缺省值，就能一边输入一边查看，当char c为缺省值时，直接输出这个节点的地址，甚至能在输入完毕之前就得到结果

#include <iostream>
#include <iomanip>
using namespace std;
struct Node{char c;int next;bool flag;
} node[100005];int main()
{int add1,add2,n;cin>>add1>>add2>>n;int common = -1;for(int i=0;i<n;i++){int add,nex;char letter;cin>>add>>letter>>nex;node[add].c = letter;node[add].next = nex;node[add].flag = false;}for(int i=add1;i!=-1;){//cout<<i<<" "<<node[i].c<<" "<<node[i].next<<endl;node[i].flag = true;i = node[i].next;}for(int i=add2;i!=-1;){if(node[i].flag){common = i;break;}i = node[i].next;}if(common != -1)cout<<setw(5)<<setfill('0')<<common<<endl;elsecout<<common<<endl;
}