PAT甲级真题及训练集(16)--1032. Sharing (25)

1032. Sharing (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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}, andNext 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 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

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提交代码

/**作者：一叶扁舟时间：16:58 2017/7/1思路：*/#include <stdio.h>#include <stdlib.h>#define SIZE 100001typedef struct SList{//char data;//数据域，如果添加这个我的vs2013编译器内存溢出，因为100001* 8，反正又没用到这个数据，就注释掉了int next;//下一个数据的地址，对应着数组的小标bool flag;//标记是否在链表中出现过}SList;int main(){SList sList[SIZE];int address, next;//char data;int N;//要输入的数据的长度int firstAddress;//链表开始的首地址jisuanint secondAddress;//第二个链表的开始地scanf("%d %d %d", &firstAddress, &secondAddress, &N);//1.获取输入数据，地址对应着下标for (int i = 0; i < N; i++){scanf("%d %c %d", &address, &data, &next);//sList[address].data = data;sList[address].next = next;sList[address].flag = false;}//2.将链表串起来int i = firstAddress;//首地址while (i != -1){sList[i].flag = true;//在链表1中出现过if (sList[i].next == -1){//说明到链表末尾break;}i = sList[i].next;}//查询链表2中和链表1中重复的结点int second = secondAddress;//首地址while (second != -1){if (sList[second].flag == true){printf("%05d", second);return 0;}if (sList[second].next == -1){//说明到链表末尾printf("-1");return 0;}second = sList[second].next;}//注意,如果不在最后面添加这个判断，pat总是有一个无法通过，可能存在一组第一个就是空的吧(其实我添加了一个return 0足以解决后面的输出问题//事实上不管用，只能猜测second 一开始就为-1,即第二个链表为空)if (second != -1){printf("%05d", second);}else{printf("-1");}system("pause");return 0;}