【PAT甲级】Sharing
- 2019 年 11 月 8 日
- 筆記
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本文链接：https://blog.csdn.net/weixin_42449444/article/details/89674601
Problem Description：
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:
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：
我一开始的思路是：先用一个map来记录每个字符所对应的结点地址(map的key是字符，map的value是结点所在地址)，然后按照顺序把结点存放在2个vector里，还原出俩个单词word1、word2(类型为string)。接下来就是找共同后缀啦，用ans(初始化为-1)来记录这俩个单词的共同后缀的起始结点地址，最后从后往前找出这俩个单词的共同后缀的起始结点地址进行输出即可。然而！提交代码之后有2个测试点出现了WA，这就很难受啦。果断换了一种思路，让你WA！老子生气啦！?我不还原出那2个单词啦，直接用set来记录出现第一个单词中出现过的字符所在结点地址，用flag来判断俩个单词是否有共同后缀，要是第二个单词中某个字符所在结点地址也出现在了set中果断用ans来记录这个结点地址并标记flag为true，否则flag为false输出-1。
AC代码： 20分代码：
#include <bits/stdc++.h> using namespace std; #define MAX 100005 struct Node { int address; //当前结点的地址 char data; //当前结点的数据 int next; //下一结点的地址 }node[MAX]; int main() { int Head1,Head2,N; cin >> Head1 >> Head2 >> N; map<char, int> m; //map用来记录每个字符所对应的结点地址 for(int i = 0; i < N; i++) { int temp; cin >> temp; node[temp].address = temp; cin >> node[temp].data >> node[temp].next; m[node[temp].data] = temp; } vector<Node> v1,v2; //按照顺序来存放结点 string word1 = "", word2 = ""; //第一个单词word1,第二个单词word2 for(int i = Head1; i != -1; i = node[i].next) { v1.push_back(node[i]); word1 += node[i].data; } for (int i = Head2; i != -1; i = node[i].next) { v2.push_back(node[i]); word2 += node[i].data; } // cout << word1 << " " << word2 << endl; bool flag = false; //判断word1、word2有没有共同后缀 int ans = -1; //用来记录word1、word2共同后缀的起始结点地址 for(int i = word1.length()-1, j = word2.length()-1; word1[i] == word2[j]; i--, j--) { flag = true; ans = m[word1[i]]; } if(flag) //若word1、word2有共同后缀 { printf("%05dn", ans); } else //若没有共同后缀,输出-1。-1是不需要前置补0的 { cout << ans << endl; } return 0; }
AC代码：
#include <bits/stdc++.h> using namespace std; #define MAX 100005 struct Node { int address; //当前结点的地址 char data; //当前结点的数据 int next; //下一结点的地址 }node[MAX]; int main() { int Head1,Head2,N; cin >> Head1 >> Head2 >> N; for(int i = 0; i < N; i++) { int temp; cin >> temp; node[temp].address = temp; cin >> node[temp].data >> node[temp].next; } bool flag = false; //判断word1、word2有没有共同后缀 int ans = -1; //用来记录word1、word2共同后缀的起始结点地址 set<int> s; //用来记录word1中出现过的字母所在结点地址 vector<Node> v; //按照顺序来存放word1中的结点 for(int i = Head1; i != -1; i = node[i].next) { v.push_back(node[i]); s.insert(node[i].address); } for (int i = Head2; i != -1; i = node[i].next) { if(s.count(node[i].address) != 0) { flag = true; //说明word1、word2存在共同后缀 ans = node[i].address; break; } } if(flag) //若word1、word2有共同后缀 { printf("%05dn", ans); } else //若没有共同后缀,输出-1。-1是不需要前置补0的 { cout << ans << endl; } return 0; }
