PAT 1032 Sharing [静态链表]

最新推荐文章于 2024-11-16 16:44:44 发布

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最新推荐文章于 2024-11-16 16:44:44 发布

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本文介绍了一种使用链表存储英语单词并寻找两个单词公共后缀的算法。通过存储单词的每个字母并让具有相同后缀的单词共享子链表，节省了存储空间。文章详细解释了输入输出规范，提供了示例并展示了如何通过标记节点来高效查找公共后缀的起始位置。

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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:

Sample Input 2:

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

Sample Output 2:

-1

-------------------------------------这是题目和解题的分割线-------------------------------------

当前地址可以作为坐标，数据和指向地址建结构体存储。同时结构体设置flag，遍历其中一个地址时，将其全部置为1,再在遍历另一个地址时寻找第一个flag=1的地址。

注意：不是找第一个字母相同的地址，而是第一个地址相同的地址。所以只要出现了相同地址，那么接下来的走向（字母）是一模一样的。

#include<cstdio>

struct
{
	char data;
	int next;
	int address;
	int flag;
}node[100010];

int main()
{
	int a,b,n,i;
	int address,nextA;
	char x;
	scanf("%d%d%d",&a,&b,&n);
	for(i=0;i<n;i++)
	{
		//有字符读入需加空格，不然会误读 
		scanf("%d %c %d",&address,&x,&nextA);
		node[address].next = nextA; //当前地址作为下标 
		node[address].data = x;
		node[address].flag = 0;
	}
	int newA = a,newB = b;
	//遍历其中一个地址，并置flag为1 
	//newA!=1不是newA.next!=1，因为循环里不断将newA.next赋给newA 
	while(newA!=-1)
	{
		node[newA].flag = 1;
		newA = node[newA].next;
	}
	//找到第一个出现的地址，中断循环 
	while(newB!=-1)
	{
		if(node[newB].flag==1) break;
		newB = node[newB].next;
	}
	if(newB==-1) printf("-1\n");
	//输出%05d 
	else printf("%05d\n",newB);
	return 0; 
}