1032. Sharing (25)
2016-02-22 17:26
591 查看
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
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:
Sample Output 1:
Sample Input 2:
Sample Output 2:
#include <iostream>
#include <cstdio>
#include <unordered_map>
using namespace std;
struct Node{
int next;
bool used;
Node(int n) : next(n), used(false){}
};
int main(){
int node1, node2, n;
scanf("%d%d%d", &node1, &node2, &n);
unordered_map<int, Node*> nodes;
for(int i = 0; i < n; ++i){
int address, next;
char data;
scanf("%d %c %d", &address, &data, &next);
nodes[address] = new Node(next);
}
while(node1 != -1){
nodes[node1]->used = true;
node1 = nodes[node1]->next;
}
while(node2 != -1){
if(nodes[node2]->used){
printf("%05d", node2);
return 0;
}
node2 = nodes[node2]->next;
}
printf("-1");
return 0;
}
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
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:
67890
Sample Input 2:
00001 00002 4 00001 a 10001 10001 s -1 00002 a 10002 10002 t -1
Sample Output 2:
-1
#include <iostream>
#include <cstdio>
#include <unordered_map>
using namespace std;
struct Node{
int next;
bool used;
Node(int n) : next(n), used(false){}
};
int main(){
int node1, node2, n;
scanf("%d%d%d", &node1, &node2, &n);
unordered_map<int, Node*> nodes;
for(int i = 0; i < n; ++i){
int address, next;
char data;
scanf("%d %c %d", &address, &data, &next);
nodes[address] = new Node(next);
}
while(node1 != -1){
nodes[node1]->used = true;
node1 = nodes[node1]->next;
}
while(node2 != -1){
if(nodes[node2]->used){
printf("%05d", node2);
return 0;
}
node2 = nodes[node2]->next;
}
printf("-1");
return 0;
}
相关文章推荐
- 使用C++实现JNI接口需要注意的事项
- 关于指针的一些事情
- c++ primer 第五版 笔记前言
- share_ptr的几个注意点
- Lua中调用C++函数示例
- Lua教程(一):在C++中嵌入Lua脚本
- Lua教程(二):C++和Lua相互传递数据示例
- C++联合体转换成C#结构的实现方法
- C++高级程序员成长之路
- C++编写简单的打靶游戏
- C++ 自定义控件的移植问题
- C++变位词问题分析
- C/C++数据对齐详细解析
- C++基于栈实现铁轨问题
- C++中引用的使用总结
- 使用Lua来扩展C++程序的方法
- C++中调用Lua函数实例
- Lua和C++的通信流程代码实例
- C与C++之间相互调用实例方法讲解
- 解析C++中派生的概念以及派生类成员的访问属性