(PAT)1032 Sharing (可以用数组表示地址)

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 beingare 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:

67890

Sample Input 2:

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

Sample Output 2:

-1

解题思路:

我们可以把数组大小定义成100010,这样就形成了一个地址池,根据题目中给出的自身地址和next地址,可以利用这个地址池进行存储.寻找第一个公共节点时,我们可以让链表1先进行遍历,将遍历到的节点都标记位true

然后让链表2进行遍历,遍历到为true的节点就是第一个公共节点

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100010;  //地址池

class Node {
public:
	int next;
	char algpa;
	bool flag;
}node[maxn];

int main() {
	for (int i = 0; i < maxn; ++i) {
		node[i].flag = false;
	}

	int firstAddr = 0;
	int secondAddr = 0;
	int n = 0;
	cin >> firstAddr >> secondAddr >> n;

	for (int i = 0; i < n; ++i) {
		int addr,next = 0;
		char data;
		cin >> addr >> data >> next;
		node[addr].next = next;
		node[addr].algpa = data; 
	}

	int p1 = firstAddr;  //第一个首地址
	int p2 = secondAddr; //第二个首地址

	while (p1 != -1) {  //先把第一条链表做上标记
		node[p1].flag = true;
		p1 = node[p1].next;   //向下遍历
	}

	while (p2 != -1) {
		if (node[p2].flag == true) {  //找到第一个重合词
			break;
		}
		p2 = node[p2].next;
	}

	if (p2 != -1) {
		printf("%05d\n", p2);
	}
	else {
		printf("-1\n");
	}

	return 0;
}