图论练习6

[NOIP2013]车站分级

Here

解题思路

由于起始点之间所选的站号，相互之间一定满足 $\geq$
那么对于起始点间未选择的站号，一定满足 $<$ 选择的站号
考虑用边来维护 $<$ 信息， $u\rightarrow v$ 表示 $u$ 的级别大于 $v$
按题意，则车站会被分为几个联通块，且保证块内无环，即一定存在一个有序序列
则通过拓扑排序，找到连通块内的最长路径，即为最小的必要分级数


import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.BitSet;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.Objects;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.Vector;









public class Main{
	static long md=(long)998244353;
	static long Linf=Long.MAX_VALUE/2;
	static int inf=Integer.MAX_VALUE/2;


	public static void main(String[] args) throws IOException{
		AReader input=new AReader();
	    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));	
	    /*
	     * 用有向边维护大于号
	     * 按题意，则车站会被分为几个联通块，且保证块内无环，即一定存在一个有序序列
	     * 最长的序列即为最小的必要分级数
	    */
	    
	    int n=input.nextInt();
	    int m=input.nextInt();
	    boolean[][] e=new boolean[n+1][n+1];
	    boolean[] vis=new boolean[n+1];
	    int[] st=new int[n+1];
	    int[] in=new int[n+1];
	    for(int i=1;i<=m;++i) {
	    	int t=input.nextInt();
	    	Arrays.fill(vis, false);
	    	for(int j=1;j<=t;++j) {
	    		int x=input.nextInt();
	    		vis[x]=true;
	    		st[j]=x;
	    	}
	    	for(int j=st[1];j<=st[t];++j) {
	    		if(vis[j])continue;
	    		for(int k=1;k<=t;++k) {
	    			if(e[st[k]][j])continue;
	    			e[st[k]][j]=true;
	    			in[j]++;
	    		}
	    	}
	    }
	    Queue<Integer> q=new LinkedList<Integer>();
	    for(int i=1;i<=n;++i)if(in[i]==0)q.add(i);
	    int[] a=new int[n+1];
	    Arrays.fill(a, 1);
	    while(!q.isEmpty()) {
	    	int x=q.peek();
	    	q.poll();
	    	for(int i=1;i<=n;++i) {
	    		if(e[x][i]) {
	    			a[i]=Math.max(a[i], a[x]+1);
	    			in[i]--;
	    			if(in[i]==0)q.add(i);
	    		}
	    		
	    	}
	    }
	    int mx=0;
	    for(int i=1;i<=n;++i)mx=Math.max(mx, a[i]);
	    out.print(mx);
 	    out.flush();
	    out.close();
	}
	//System.out.println();
	//out.println();
	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();
	    }
	}
}

P4366 [Code+#4] 最短路

Here

解题思路

主要是对于 $i\rightarrow j,w=(i\ xor j)*c$ ，如何处理，避免建 $n(n-1)$ 条边
考虑二进制
$i\rightarrow j$ ，可以看作是通过某些 $w=2^k*c$ 的边到达
设 $k=2^k$ ，则 $i\rightarrow (i\ xor \ k),w=k,(i\ xor\ k)\leq n$ ，建边
然后跑最短路即可
注意 $3\rightarrow 4\Rightarrow 3\overset{1*c}{\rightarrow}2\overset{2*c}{\rightarrow}0\overset{4*c}{\rightarrow}4$ ，需要添一点0
边总共有 $m+n*20$
洛谷该题内存不够，可以点链接去牛客交


import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.PriorityQueue;
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=500010+20*100010;
	static int n=0;
	static int m=0;
	static int c=0;
	
	static
	class Edge{
		int fr,to,nxt;
		int val;
		public Edge(int u,int v,int w) {
			fr=u;
			to=v;
			val=w;
		}
	}
	static Edge[] e;
	static int[] head;
	static int cnt=0;
	static void addEdge(int fr,int to,int val) {
		cnt++;
		e[cnt]=new Edge(fr, to, val);
		e[cnt].nxt=head[fr];
		head[fr]=cnt;
	}
	static
	class Node{
		int x;
		int y;
		public Node(int u,int v) {
			x=u;
			y=v;
		}
	}
	static int Dij(int s,int t) {
		int[] dis=new int[n+1];
		Arrays.fill(dis, inf);
		boolean[] vis=new boolean[n+1];
		dis[s]=0;
		PriorityQueue<Node> q=new PriorityQueue<Node>((o1,o2)->{
			return o1.y-o2.y;
		});
		q.add(new Node(s, 0));
		while(!q.isEmpty()) {
			Node now=q.peek();
			q.poll();
			int x=now.x;
			if(vis[x])continue;
			vis[x]=true;
			for(int i=head[x];i>0;i=e[i].nxt) {
				int v=e[i].to;
				int w=e[i].val;
				if(vis[v])continue;
				if(dis[v]>dis[x]+w) {
					dis[v]=dis[x]+w;
					q.add(new Node(v, dis[v]));
				}
			}
		}
		return dis[t];
	}
	static void solve() throws Exception{
		AReader input=new AReader();
	    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));	
	    n=input.nextInt();
	    m=input.nextInt();
	    c=input.nextInt();
	    e=new Edge[m+20*n+1];
	    head=new int[n+1];
	    cnt=0;
	    for(int i=1;i<=m;++i) {
	    	int u=input.nextInt();
	    	int v=input.nextInt();
	    	int w=input.nextInt();
	    	
	    	addEdge(u, v, w);
	    }
	    for(int i=0;i<=n;++i) {
	    	for(int j=1;j<=n;j<<=1) {
	    		int x=i^j;
	    		if(x>n)continue;
	    		addEdge(i, x, j*c);
	    	}
	    }
	    int s=input.nextInt();
	    int t=input.nextInt();
	    out.print(Dij(s, t));
 	    out.flush();
	    out.close();
	}
	public static void main(String[] args) throws Exception{
		solve();
	}
//	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();
		    }
		}
	}