1032 Sharing （附个人吐槽）

1032 Sharing （25 分）

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 -1instead.

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

1032 Sharing （附个人吐槽）

吐槽：pat数据漏洞较大，一开始想简单了，发现数据和我一样想简单了。首先，一个单词内部如果有相同的字母这个测试点没有考虑到，我虽然对两个单词内字母出现分别加100和加1处理，但是判断的时候判101，还是考虑没有重复字母；其次，字母的值可以忽略不管；问题只需要判断开始的结点地址是否相同，即使两个单词内有一个相同的字母他也认为是相同后缀。

#include<iostream>
using namespace std;
int Mcount[100000] = { 0 },words[100000],s1,s2,n,t,f=1,d;
char c;

int main() {
	cin >> s1 >> s2 >> n;
	for (int i = 0; i < n; i++) {
		scanf("%d %c %d", &t, &c, &d);
		words[t] = d;
	}
	while (s1 != -1 || s2 != -1) {
		Mcount[s1]++;
		Mcount[s2]+=100;
		if (Mcount[s2] == 101) {
			printf("%05d", s2);
			return 0;
		}
		else if (Mcount[s1] == 101) {
			cout<<s1; return 0;
		}
		if(s1!=-1)s1 = words[s1];
		if (s2 != -1)s2 = words[s2];
	}
	if (f)cout << "-1";
	return 0;
}