1032. Sharing (25)

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

where Address 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 Nextis 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 967890 i 0000200010 a 1234500003 g -112345 D 6789000002 n 0000322222 B 2345611111 L 0000123456 e 6789000001 o 00010

Sample Output 1:

67890

Sample Input 2:

00001 00002 400001 a 1000110001 s -100002 a 1000210002 t -1

Sample Output 2:

-1

这题就是找相同后缀的地址

其实就是看两个单词同时出现的第一个地址是什么，后面地址肯定是一样的

假如第一个单词的地址已经确定了，第二个单词只要出现第一个单词的某个地址，这个地址就是输出的结果

举个例子：

(1)l(2)o(3)a(4)i(5)n(6)g

第一个单词的地址都建立好了

第二个单词 

boing 假设b的地址是7,o的地址已经确定了，o的下一个地址也是确定的，所以出现相同地址时就是输出的结果

我第一次用的while循环。。。超时一分，改的for过了，不知为啥

#include<iostream>  #include<cstring>  #include<cstdio>  #include<queue>  #include<stack>  #include<algorithm>  #include<vector> #include<set>using namespace std;typedef struct node{int next;char data;}node;node no[100000];int main(){int q1,q2,n;cin>>q1>>q2>>n;for(int i=0;i<n;i++){int a,c;char b;scanf("%d %c %d",&a,&b,&c);no[a].data=b;no[a].next=c;}int v[100000]={0};/*while(1){v[q1]=1;q1=no[q1].next;if(q1==-1) break;}*/    for(;q1!=-1;q1=no[q1].next) v[q1]=1;/*while(1){if(v[q2]==1){printf("%05d",q2);return 0;}q2=no[q2].next;if(q2==-1) break; }*/for(;q2!=-1;q2=no[q2].next){if(v[q2]==1) {printf("%05d",q2);return 0;}}cout<<-1;return 0;}