拓扑-士兵排队问题2

╰+攻爆jí腚メ 2022-07-12 12:15 130阅读 0赞

package 搜索.拓扑;  
import java.util.Collections;  
import java.util.HashMap;  
import java.util.LinkedList;  
import java.util.Map;  
import java.util.Queue;  
import java.util.Scanner;  
import java.util.Set;  
  
  
/\*  
\* 使用一键多值，不需要去重，因为，集合里面保存了重复信息  
\* 要解决排序问题。因为是从前往后排序的  
\* 排序问题  
\* map\#LinkedList进行排序  
\* 去重问题  
\* 不去重，不过要排序  
\* 去重(前提，判断是否有，有n【i】则不加1)：  
\* hash结构|数组  
\*   
\*/  
public class 士兵排队问题2 \{  
static Queue<Integer> que = new LinkedList<Integer>();  
static Map<Integer, LinkedList<Integer>> map = new HashMap<>();  
  
  
public static void main(String\[\] args) \{  
Scanner sc = new Scanner(System.in);  
int n = Integer.parseInt(sc.nextLine());  
char result\[\] = new char\[26\];  
int count = 0;  
int\[\] in = new int\[26\];  
boolean visited\[\] = new boolean\[26\];  
for (int i = 0; i < n; i++) \{  
char\[\] charArray = sc.nextLine().toCharArray();  
int a = charArray\[0\] - 'A';  
int b = charArray\[2\] - 'A';  
LinkedList<Integer> linkedlist = map.get(a);  
if (linkedlist == null) \{  
linkedlist = new LinkedList<>();  
\}  
linkedlist.add(b);  
map.put(a, linkedlist);  
in\[b\]++;  
visited\[a\] = true;  
visited\[b\] = true;  
\}  
Set<Integer> keySet = map.keySet();  
for (Integer i : keySet) \{  
LinkedList<Integer> linkedList = map.get(i);  
Collections.sort(linkedList);  
map.put(i, linkedList);  
\}  
int size = 0;  
for (int i = 0; i < 26; i++) \{  
if (visited\[i\]) \{  
size++;  
if (in\[i\] == 0)  
que.add(i);  
\}  
\}  
while (!que.isEmpty()) \{  
Integer remove = que.remove();  
result\[count++\] = (char) ('A' + remove);  
LinkedList<Integer> linkedlist = map.get(remove);  
if (linkedlist == null)  
continue;  
for (int i : linkedlist) \{  
in\[i\]--;  
if (in\[i\] == 0)  
que.add(i);  
\}  
\}  
if (count != size)  
System.out.println("No Answer!");  
else  
for (int i = 0; i < size; i++) \{  
System.out.print(result\[i\]);  
\}  
\}  
\}