牛客小白月赛88

E.多重映射

解题思路
  • 对集合进行整体操作,集合大小只增不减,问最后集合标号
  • 维护集合,考虑并查集
  • 但直接用并差集维护会有以下问题:
  • 当前集合变标号,可能会和之前标号相同,则进行并查集find操作时,会接上之前的变换
  • 为消除其影响,考虑加入时间戳,后面新产生的标号与之前的不同
  • 进行操作时,若x没有则跳过
  • y没有,则产生新的y集合,y对应的时间戳++
  • 若有,则合并集合,时间戳不变
  • fa[Node(x,tim_x)]=Node(y,tim_y)
  • find时,带上时间戳
import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Objects;
import java.util.PriorityQueue;
import java.util.Stack;
import java.util.StringTokenizer;








//implements Runnable
public class Main{
	static long md=(long)998244353;
	static long Linf=Long.MAX_VALUE/2;
	static int inf=Integer.MAX_VALUE/2;
	static int N=2000000;
	static int n=0;
	static int m=0;
	
	static 
	class Node{
		int x;
		int y;
		public Node(int u,int v) {
			x=u;
			y=v;
		}
		@Override
	    public boolean equals(Object o) {
	        if (this == o) return true;
	        if (o == null || getClass() != o.getClass()) return false;
	        Node may = (Node) o;
	        return x == may.x && y==may.y;
	    }

	    @Override
	    public int hashCode() {
	        return Objects.hash(x, y);
	    }
	}
	static HashMap<Node, Node> fa=new HashMap<Node, Node>();
	static Node find(Node x) {//跟正常并查集一样
		if(x.x==fa.get(x).x&&x.y==fa.get(x).y)return x;
		else {
			Node f=fa.get(x);
			Node ff=find(f);
			fa.put(f, ff);
			return ff;
		}
	}
	static int[] a=new int[N+1];
	static int[] x=new int[N+1];
	static int[] y=new int[N+1];
	static boolean[] hs=new boolean[N+1];
	static int[] t=new int[N+1];
	public static void main(String[] args) throws Exception{
		AReader input=new AReader();
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));	
		int T=input.nextInt();
		while(T>0) {
			n=input.nextInt();
			m=input.nextInt();
			
			for(int i=1;i<=n;++i) {
				a[i]=input.nextInt();
                Node ro=new Node(a[i], 0);
				fa.put(ro, ro);//初始化,重复添无影响
				hs[a[i]]=true;
			}
			
			for(int i=1;i<=m;++i) {
				x[i]=input.nextInt();
				y[i]=input.nextInt();
				Node lo=new Node(x[i], 0);
				Node ro=new Node(y[i], 0);
                fa.put(lo, lo);
                fa.put(ro, ro);//初始化,重复添无影响
			}
//			out.print(fa.get(new Node(1, 0)));
			for(int i=1;i<=m;++i) {
				int l=x[i];
				int r=y[i];
				if(!hs[l])continue;
				
				if(!hs[r]) {//与之前区别
					t[r]++;
				}
                hs[l]=false;
				hs[r]=true;
				Node lo=new Node(l, t[l]);
				Node ro=new Node(r, t[r]);
				fa.put(lo, ro);
//				out.print(fa.get(ro));
				fa.put(ro, ro);//集合头自己指自己,初始化,重复添无影响
			}
			
//			out.print(fa.get(new Node(1, 0)));
			for(int i=1;i<=n;++i) {
				Node fx=find(new Node(a[i], 0));
				out.print(fx.x+" ");
			}
		    out.println();
            //清空
		    for(int i=1;i<=n;++i) {
		    	hs[a[i]]=false;
		    	t[a[i]]=0;
		    }
		    for(int i=1;i<=m;++i) {
		    	hs[x[i]]=false;
		    	t[x[i]]=0;
		    	hs[y[i]]=false;
		    	t[y[i]]=0;
		    }
            fa.clear();
			T--;
		}
	    out.flush();
	    out.close();
	}
//	public static final void main(String[] args) throws Exception {
//		  new Thread(null, new Main(), "线程名字", 1 << 27).start();
//	}
//		@Override
//		public void run() {
//			try {
//				//原本main函数的内容
//				solve();
//
//			} catch (Exception e) {
//			}
//		}
		static
		class AReader{ 
		    BufferedReader bf;
		    StringTokenizer st;
		    BufferedWriter bw;

		    public AReader(){
		        bf=new BufferedReader(new InputStreamReader(System.in));
		        st=new StringTokenizer("");
		        bw=new BufferedWriter(new OutputStreamWriter(System.out));
		    }
		    public String nextLine() throws IOException{
		        return bf.readLine();
		    }
		    public String next() throws IOException{
		        while(!st.hasMoreTokens()){
		            st=new StringTokenizer(bf.readLine());
		        }
		        return st.nextToken();
		    }
		    public char nextChar() throws IOException{
		        //确定下一个token只有一个字符的时候再用
		        return next().charAt(0);
		    }
		    public int nextInt() throws IOException{
		        return Integer.parseInt(next());
		    }
		    public long nextLong() throws IOException{
		        return Long.parseLong(next());
		    }
		    public double nextDouble() throws IOException{
		        return Double.parseDouble(next());
		    }
		    public float nextFloat() throws IOException{
		        return Float.parseFloat(next());
		    }
		    public byte nextByte() throws IOException{
		        return Byte.parseByte(next());
		    }
		    public short nextShort() throws IOException{
		        return Short.parseShort(next());
		    }
		    public BigInteger nextBigInteger() throws IOException{
		        return new BigInteger(next());
		    }
		    public void println() throws IOException {
		        bw.newLine();
		    }
		    public void println(int[] arr) throws IOException{
		        for (int value : arr) {
		            bw.write(value + " ");
		        }
		        println();
		    }
		    public void println(int l, int r, int[] arr) throws IOException{
		        for (int i = l; i <= r; i ++) {
		            bw.write(arr[i] + " ");
		        }
		        println();
		    }
		    public void println(int a) throws IOException{
		        bw.write(String.valueOf(a));
		        bw.newLine();
		    }
		    public void print(int a) throws IOException{
		        bw.write(String.valueOf(a));
		    }
		    public void println(String a) throws IOException{
		        bw.write(a);
		        bw.newLine();
		    }
		    public void print(String a) throws IOException{
		        bw.write(a);
		    }
		    public void println(long a) throws IOException{
		        bw.write(String.valueOf(a));
		        bw.newLine();
		    }
		    public void print(long a) throws IOException{
		        bw.write(String.valueOf(a));
		    }
		    public void println(double a) throws IOException{
		        bw.write(String.valueOf(a));
		        bw.newLine();
		    }
		    public void print(double a) throws IOException{
		        bw.write(String.valueOf(a));
		    }
		    public void print(char a) throws IOException{
		        bw.write(String.valueOf(a));
		    }
		    public void println(char a) throws IOException{
		        bw.write(String.valueOf(a));
		        bw.newLine();
		    }
		}
	}

		

	

### 关于小白109的信息 目前并未找到关于小白109的具体比信息或题解内容[^5]。然而,可以推测该事可能属于网举办的系列算法之一,通常这类比会涉及数据结构、动态规划、图论等经典算法问题。 如果要准备类似的事,可以通过分析其他场次的比题目来提升自己的能力。例如,在小白13中,有一道与二叉树相关的题目,其核心在于处理树的操作以及统计最终的结果[^3]。通过研究此类问题的解决方法,能够帮助理解如何高效地设计算法并优化时间复杂度。 以下是基于已有经验的一个通用解决方案框架用于应对类似场景下的批量更新操作: ```python class TreeNode: def __init__(self, id): self.id = id self.weight = 0 self.children = [] def build_tree(n): nodes = [TreeNode(i) for i in range(1, n + 1)] for node in nodes: if 2 * node.id <= n: node.children.append(nodes[2 * node.id - 1]) if 2 * node.id + 1 <= n: node.children.append(nodes[2 * node.id]) return nodes[0] def apply_operations(root, operations, m): from collections import defaultdict counts = defaultdict(int) def update_subtree(node, delta): stack = [node] while stack: current = stack.pop() current.weight += delta counts[current.weight] += 1 for child in current.children: stack.append(child) def exclude_subtree(node, total_nodes, delta): nonlocal root stack = [(root, False)] # (current_node, visited) subtree_size = set() while stack: current, visited = stack.pop() if not visited and current != node: stack.append((current, True)) for child in current.children: stack.append((child, False)) elif visited or current == node: if current != node: subtree_size.add(current.id) all_ids = {i for i in range(1, total_nodes + 1)} outside_ids = all_ids.difference(subtree_size.union({node.id})) for idx in outside_ids: nodes[idx].weight += delta counts[nodes[idx].weight] += 1 global nodes nodes = {} queue = [root] while queue: curr = queue.pop(0) nodes[curr.id] = curr for c in curr.children: queue.append(c) for operation in operations: op_type, x = operation.split(' ') x = int(x) target_node = nodes.get(x, None) if not target_node: continue if op_type == '1': update_subtree(target_node, 1) elif op_type == '2' and target_node is not None: exclude_subtree(target_node, n, 1) elif op_type == '3': path_to_root = [] temp = target_node while temp: path_to_root.append(temp) if temp.id % 2 == 0: parent_id = temp.id // 2 else: parent_id = (temp.id - 1) // 2 if parent_id >= 1: temp = nodes[parent_id] else: break for p in path_to_root: p.weight += 1 counts[p.weight] += 1 elif op_type == '4': pass # Implement similarly to other cases. result = [counts[i] for i in range(m + 1)] return result ``` 上述代码片段展示了针对特定类型的树形结构及其操作的一种实现方式。尽管它并非直接对应小白109中的具体题目,但它提供了一个可借鉴的设计思路。 ####
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