PAT甲级 1032 Sharing

题目链接

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 (≤1e5), 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.

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

这题感觉网上都写得好复杂，我心想第一个公共后缀的地址，不就是在输入的时候出现两次就可以了嘛，然后就用 mapmapmap 记录了一下，交了一发 222222 分，最后一个点错了，想了想很快就发现可能两个单词在某一个字符的地址是一样的，随后又分支出去了，所以这也很好改，只要再标记一下这个地址作为前缀出现了几次就行，也即第一个公共后缀地址的条件就是：作为前缀只出现了一次，作为后缀出现了两次，用两个 mapmapmap 标记即可，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
int main(){map<int,int>m1,m2;int a,b,x,y,n,ans=-1;char c;scanf("%d%d%d",&a,&b,&n);while(n--){scanf("%d %c %d",&x,&c,&y);m1[x]++;m2[y]++;}if(a==b) ans=a;for(auto i:m1){if(i.second==1&&m2[i.first]==2) ans=i.first;}if(ans>0) printf("%05d",ans);else printf("-1");
}