PTA 1032 Sharing

个人学习记录，代码难免不尽人意。

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 (≤10 5 ), 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

#include 
struct Node{
	int next;
	char data;
	bool flag;
}node[100010];
int main(){
   int begin1,begin2,n;
   scanf("%d %d %d",&begin1,&begin2,&n);
   for(int i=0;i<n;i++){
   	int address,next;
   	char data;
   	scanf("%d %c %d",&address,&data,&next);
   	node[address].data=data;
   	node[address].next=next;
   	node[address].flag=false;
   }
   int next=begin1;
   while(next!=-1){
   node[next].flag=true;
   	next=node[next].next;}

   int next2=begin2;
   bool flag=false;
   while(next2!=-1){
   	   if(node[next2].flag){
   	   	  flag=true;//这个地方需要在break前面填加
		break;
		  }
		else{
			next2=node[next2].next;
		}
   }
   if(flag) printf("%d",next2);
   else printf("-1");
   
}

本题不难，需要注意的是如果在循环中加入break语句，则直接跳出循环而不执行后面的语句，我犯了这个小失误。