1032 Sharing (25分)(数组链表)

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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 (≤), 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 −.

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
题目分析：对于第一个链表 把他们都置为访问过 在第二个链表中查找访问过的节点 (我怎么就想不出这样的解法呢。。。。)

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <climits>
 3 #include<iostream>
 4 #include<vector>
 5 #include<queue>
 6 #include<stack>
 7 #include<algorithm>
 8 #include<string>
 9 #include<cmath>
10 using namespace std;
11 struct Node{
12     char data;
13     int next;
14     bool flag;
15 }Nodes[100000];
16 int main()
17 {
18     int add1, add2,n,a,b;
19     char c;
20     cin >> add1 >> add2 >> n;
21     for (int i = 0; i < n; i++)
22     {
23         cin >> a >> c >> b;
24         Nodes[a] = { c,b,false };
25     }
26     for (int i = add1; i != -1; i = Nodes[i].next)
27         Nodes[i].flag = true;
28     for(int i=add2;i!=-1;i=Nodes[i].next)
29         if (Nodes[i].flag == true)
30         {
31             printf("%05d", i);
32             return 0;
33         }
34     cout << -1;
35     return 0;
36 }

View Code

1032 Sharing (25分)(数组链表)

标签：eve   ati   orm   ges   als   clu   eth   positive   false   

原文地址：https://www.cnblogs.com/57one/p/12006192.html