Zhejiang university----Sharing

解析英文单词存储与共享技术

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本文探讨了如何使用链表存储英文单词，并通过共享相同后缀来节省空间。介绍了如何找到公共后缀的起始位置，涉及输入解析、节点遍历及输出处理等关键步骤。

//Zhejiang university----Sharing
/*
时间限制 1s 内存限制128M

题目描述：

    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).

输入：

    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

    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 Next is the position of the next node.

输出：

    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.

样例输入：

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

样例输出：

    67890
    -1
*/
#include <stdio.h>
#include <string.h>
const int MAX = 100010;
int hash[MAX];
int mark[MAX];
int main(){
   int a,b,c,d,n,i;
   char e;
   //freopen("C:\\1.txt","r",stdin);
   while(scanf("%d%d%d",&a,&b,&n) != EOF){
       int start1 = a;
       int start2 = b;
       for(i=0;i<n;i++){
           scanf("%d %c %d",&c,&e,&d);
           hash[c] = d;
       }
       for(i=0;i<MAX;i++)
           mark[i]=0;
       int id1 = start1,id2 = start2;
       while(id1 != -1){
           mark[id1] = 1;
           id1 = hash[id1];
       }
       while(id2 != -1){
           if(mark[id2] == 1){
               printf("%05d\n",id2);
               break;
           }else
               id2 = hash[id2];
       }
       if(id2 == -1)
           printf("-1\n");
   }
   return 0;
}