题目1468: Sharing
时间限制：1 秒
内存限制：128 兆
特殊判题：否
提交：367
解决：28
- 题目描述：
-
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).
- 输入：
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For each case, the first line contains two addresses of nodes and a positive N (<= 10^5), 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.
- 输出：
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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.
- 样例输入：
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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 00001 00002 4 00001 a 10001 10001 s -1 00002 a 10002 10002 t -1
- 样例输出：
-
67890 -1
- 来源：
- 2012年浙江大学计算机及软件工程研究生机试真题
1 #include<iostream> 2 #include<string> 3 #include<map> 4 #include<stdio.h> 5 using namespace std; 6 struct node{ 7 int address; 8 int flag; 9 int next; 10 }; 11 map<int,node*> m; 12 node* getNode(int address){ 13 return (node*)(*(m.find(address))).second; 14 } 15 int main(){ 16 int addsA,addsB; 17 int node_num; 18 while(cin>>addsA>>addsB>>node_num){ 19 for(node_num;node_num>0;node_num--){ 20 node *p = new node; 21 char data; 22 cin>>p->address>>data>>p->next; 23 p->flag = 0; 24 m[p->address] = p; 25 } 26 int j = addsA; 27 while(j != -1){ 28 getNode(j)->flag += 1; 29 j = getNode(j)->next; 30 31 } 32 j = addsB; 33 int start = -1; 34 while(j != -1){ 35 (getNode(j)->flag) += 1; 36 if(getNode(j)->flag == 2){ 37 start = getNode(j)->address; 38 break; 39 } 40 j = getNode(j)->next; 41 } 42 if(start != -1) printf( "%.05d\n",start); 43 else printf("-1\n"); 44 map<int,node*>::iterator itr = m.begin(); 45 while(itr != m.end()){ 46 delete (*itr).second; 47 itr++; 48 } 49 m.clear(); 50 } 51 return 0; 52 }
