0
这是我问here的问题的延续。
鉴于节点(雇员)总数和邻接表(员工之间的友谊),我需要找到所有连接的组件
。下面是我的代码 -BFS实现查找连接组件时间过长
public class Main {
static HashMap<String, Set<String>> friendShips;
public static void main(String[] args) throws IOException {
BufferedReader in= new BufferedReader(new InputStreamReader(System.in));
String dataLine = in.readLine();
String[] lineParts = dataLine.split(" ");
int employeeCount = Integer.parseInt(lineParts[0]);
int friendShipCount = Integer.parseInt(lineParts[1]);
friendShips = new HashMap<String, Set<String>>();
for (int i = 0; i < friendShipCount; i++) {
String friendShipLine = in.readLine();
String[] friendParts = friendShipLine.split(" ");
mapFriends(friendParts[0], friendParts[1], friendShips);
mapFriends(friendParts[1], friendParts[0], friendShips);
}
Set<String> employees = new HashSet<String>();
for (int i = 1; i <= employeeCount; i++) {
employees.add(Integer.toString(i));
}
Vector<Set<String>> friendBuckets = bucketizeEmployees(employees);
System.out.println(friendBuckets.size());
}
public static void mapFriends(String friendA, String friendB, Map<String, Set<String>> friendsShipMap) {
if (friendsShipMap.containsKey(friendA)) {
friendsShipMap.get(friendA).add(friendB);
} else {
Set<String> friends = new HashSet<String>();
friends.add(friendB);
friendsShipMap.put(friendA, friends);
}
}
public static Vector<Set<String>> bucketizeEmployees(Set<String> employees) {
Vector<Set<String>> friendBuckets = new Vector<Set<String>>();
while (!employees.isEmpty()) {
String employee = getHeadElement(employees);
Set<String> connectedEmployeesBucket = getConnectedFriends(employee);
friendBuckets.add(connectedEmployeesBucket);
employees.removeAll(connectedEmployeesBucket);
}
return friendBuckets;
}
private static Set<String> getConnectedFriends(String friend) {
Set<String> connectedFriends = new HashSet<String>();
connectedFriends.add(friend);
Set<String> queuedFriends = new LinkedHashSet<String>();
if (friendShips.get(friend) != null) {
queuedFriends.addAll(friendShips.get(friend));
}
while (!queuedFriends.isEmpty()) {
String poppedFriend = getHeadElement(queuedFriends);
connectedFriends.add(poppedFriend);
if (friendShips.containsKey(poppedFriend))
for (String directFriend : friendShips.get(poppedFriend)) {
if (!connectedFriends.contains(directFriend) && !queuedFriends.contains(directFriend)) {
queuedFriends.add(directFriend);
}
}
}
return connectedFriends;
}
private static String getHeadElement(Set<String> setFriends) {
Iterator<String> iter = setFriends.iterator();
String head = iter.next();
iter.remove();
return head;
}
}
我曾尝试使用下面的脚本测试我的代码,其结果我消耗为sdtIn -
#!/bin/bash
echo "100000 100000"
for i in {1..100000}
do
r1=$(($RANDOM % 100000))
r2=$(($RANDOM % 100000))
echo "$r1 $r2"
done
,而我是能够验证(为小事输入),我的答案是正确的,当我尝试与上面的脚本巨大的投入,我看到运行需要很长时间(〜20s)。
任何我可以在我的实施中做得更好的东西?
尝试使用ArrayList代替运行仍需要约20秒 – ping 2013-03-09 19:36:34
可能它读取文件首先读取文件并排除时间测量 – AlexWien 2013-03-09 19:41:18
排除读取部分,执行需要18002女士。 – ping 2013-03-09 20:07:47