題目

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

剛看了靜態鏈表這一節，挺好，讓我寫的話估計還想著用普通鏈表記錄地址，這樣再去掃描比較

這里很重要的思路就是一個節點只會有一個next，即使有多個pre。

也就是說，只要確定兩條鏈表同時出現的第一個節點，它后面的內容就一定一樣

結構體數組沒法直接初始化（要初始化得遍歷數組），不像類那樣直接來個默認值，所以一次遍歷鏈表是不可能完成的。

如果用類的話帶個缺省值，就能一邊輸入一邊查看，當char c為缺省值時，直接輸出這個節點的地址，甚至能在輸入完畢之前就得到結果

#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;
}