图论练习1

内容： $Tuopu$ ，拆点，分层， $Bitset$ 传递，带限制的最小生成树

[HNOI2015]菜肴制作

题目链接

题目大意

有 $m$ 个限制， $x$ 号菜肴在 $y$ 号前完成
在满足限制的条件下，按照 $1\cdots 2\cdots 3\cdots$ 出菜( $1\cdots 2$ 是为了满足 $2$ 的限制 )

解题思路

由限制，可以考虑 $Tuopu$
若直接正向 $Tuopu$ ，以 $3\rightarrow 4\rightarrow 1 \ | \ 2 \ | \ 5$ 为例，则会先出 $2$
而反向 $Tuopu$ ， $1\rightarrow 4\rightarrow 3 \ | \ 2 \ | \ 5$
此时对于一路限制，最先出的最小的号
题目有要求先满足较小号的限制
所以将队列改为由大到小排序的堆，再倒序输出每次出堆的号
排序的内容实际为正向限制路径上的最终菜肴
有环则无解


import java.io.*;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;




public class Main{
	static int cnt=0;
	static Edge[] e;
	static int[] head;
	static
	class Edge{
		long val;
		int to;
		int fr;
		int nxt;
	}
	static void addEdge(int fr,int to,long val) {
		cnt++;
		e[cnt]=new Edge();
		e[cnt].fr=fr;
		e[cnt].to=to;
		e[cnt].val=val;
		e[cnt].nxt=head[fr];
		head[fr]=cnt;
	}
	
	static
	class Node{
		int x;
		long dis;
		public Node() {}
		public Node(int I,long D) {
			x=I;;
			dis=D;
		}
	}
	
	public static void main(String[] args) throws IOException{
		AReader input=new AReader();
	    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	    int T=input.nextInt();
	    while(true) {
	    	if(T==0) break;
	    	int n=input.nextInt();
		    int m=input.nextInt();
		    e=new Edge[m+1];
		    head=new int[n+1];
		    cnt=0;
		    int[] in=new int[n+1];
		    for(int i=1;i<=m;++i) {
		    	int u=input.nextInt();
		    	int v=input.nextInt();
		    	in[u]++;
		    	addEdge(v, u, 0);
		    }
		    PriorityQueue<Integer> qu=new PriorityQueue<Integer>((o1,o2)-> {return o2-o1;});
		    boolean[] vis=new boolean[n+1];
		    for(int i=1;i<=n;++i) {
		    	if(in[i]==0) {
		    		qu.add(i);
		    		vis[i]=true;
		    	}
		    }
		    int[] op=new int[n+1];
		    int cnt=0;
		    while(!qu.isEmpty()) {
		    	int x=qu.peek();
		    	qu.poll();
		    	cnt++;
		    	op[cnt]=x;
		    	for(int i=head[x];i>0;i=e[i].nxt) {
		    		int v=e[i].to;
		    		if(vis[v])continue;
		    		in[v]--;
		    		if(in[v]==0) {
		    			qu.add(v);
		    			vis[v]=true;
		    		}
		    	}
		    }
		    if(cnt!=n) {
		    	out.println("Impossible!");
		    }else {
		    	for(int i=cnt;i>0;--i) {
		    		out.print(op[i]+" ");
		    	}
		    	out.println();
		    }
	    	T--;
	    }
	    out.flush();
	    out.close();
	}
	
	static
    class AReader {
        private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        private StringTokenizer tokenizer = new StringTokenizer("");
        private String innerNextLine() {
            try {
                return reader.readLine();
            } catch (IOException ex) {
                return null;
            }
        }
        public boolean hasNext() {
            while (!tokenizer.hasMoreTokens()) {
                String nextLine = innerNextLine();
                if (nextLine == null) {
                    return false;
                }
                tokenizer = new StringTokenizer(nextLine);
            }
            return true;
        }
        public String nextLine() {
            tokenizer = new StringTokenizer("");
            return innerNextLine();
        }
        public String next() {
            hasNext();
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
        public long nextLong() {
            return Long.parseLong(next());
        }
 
    }
}

胖胖的牛牛

题目链接

题目大意

在图上从 $A$ 到 $B$ ， $"x"$ 不可经过，有水平和垂直两个方向
在 $x$ 且方向为垂直，往左或右走，需要转变方向
求 $A\rightarrow B$ 最少的转变方向次数

解题思路

将 $x\rightarrow y$ ，看作两步，先判断转向，再移动
所以将一个点拆为上下左右4个， $(u,v,l,r)$
$\left\{\begin{matrix} u\overset{0}\leftrightharpoons v\\ l\overset{0}\leftrightharpoons r\\ l\overset{1}\leftrightharpoons u\\ u\overset{1}\leftrightharpoons r\\ r\overset{1}\leftrightharpoons v\\ v\overset{1}\leftrightharpoons l \end{matrix}\right.$ ， $\left\{\begin{matrix} nxtr\overset{0}\leftrightharpoons l \\ r\overset{0}\leftrightharpoons nxtl & \\ nxtv\overset{0}\leftrightharpoons u & \\ v\overset{0}\leftrightharpoons nxtu& \end{matrix}\right.$ ， $\left\{\begin{matrix} num=(i-1)*n+j\\ l=num*4\\ u=l+1\\ r=l+2\\ v=l+3 \end{matrix}\right.$
求拆点后， $A\rightarrow B$ 的最短路


import java.io.*;
import java.io.ObjectInputStream.GetField;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.Vector;



public class Main{
	static int inf=Integer.MAX_VALUE/2;
	static
	class Edge{
		int fr,to,nxt;
		int val;
		public Edge(int u,int v,int w) {
			fr=u;
			to=v;
			val=w;
		}
	}
	static
	class Node{
		int x,y;
		public Node(int X,int Y) {
			x=X;
			y=Y;
		}
	}
	static 
	class Map{
		int[] head;
		int cnt;
		Edge[] e;
		int ans=inf;
		public Map(int n,int m) {
			e=new Edge[m];
			head=new int[n+1];
			cnt=0;
		}
		void addEdge(int fr,int to,int val) {
			cnt++;
			e[cnt]=new Edge(fr, to, val);
			e[cnt].nxt=head[fr];
			head[fr]=cnt;
		}
		void Dij(int s,int n,int[] t) {
			int[] dis=new int[n+1];
			boolean[] vis=new boolean[n+1];
			PriorityQueue<Node> q=new PriorityQueue<Node>((o1,o2)->{return o1.y-o2.y;});
			for(int i=1;i<=n;++i)dis[i]=inf;
			dis[s]=0;
			q.add(new Node(s, 0));
			while(!q.isEmpty()) {
				Node now=q.peek();
				q.poll();
				int x=now.x;
				if(vis[x])continue;
				int disu=now.y;
				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]>disu+w) {
						dis[v]=disu+w;
						q.add(new Node(v,dis[v]));
					}
				}
			}
			
			ans=Math.min(ans, Math.min(dis[t[0]], dis[t[1]]));
		}
	}
	public static void main(String[] args) throws IOException{
		AReader input=new AReader();
	    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	    int n=input.nextInt();
	    char[][] map=new char[n+1][n+1];
	    for(int i=1;i<=n;++i) {
	    	for(int j=1;j<=n;++j)map[i][j]=input.nextChar();
	    }
	    int N=4*n*n;
	    Map T=new Map(N, 200000);
	    int[] s=new int[2];
	    int[] t=new int[2];
	    for(int i=1;i<=n;++i) {
	    	for(int j=1;j<=n;++j) {
	    		if(map[i][j]=='x')continue;
	    		int num=(i-1)*n+j;
    			int l=(num-1)*4;
    			int u=l+1;
    			int r=l+2;
    			int v=l+3;
    			T.addEdge(l, u, 1);
    			T.addEdge(u, l, 1);
    			T.addEdge(u, r, 1);
    			T.addEdge(r, u, 1);
    			T.addEdge(r, v, 1);
    			T.addEdge(v, r, 1);
    			T.addEdge(v, l, 1);
    			T.addEdge(l, v, 1);
    			T.addEdge(l, r, 0);
    			T.addEdge(r, l, 0);
    			T.addEdge(u, v, 0);
    			T.addEdge(v, u, 0);
    			if(j+1<=n&&map[i][j+1]!='x') {
    				int nxtl=num*4;
    				T.addEdge(r, nxtl, 0);
    				T.addEdge(nxtl, r, 0);
    			}
    			if(j>=2&&map[i][j-1]!='x') {
    				int nxtr=(num-2)*4+2;
    				T.addEdge(nxtr, l, 0);
    				T.addEdge(l, nxtr, 0);
    			}
    			if(i>=2&&map[i-1][j]!='x') {
    				int nxtv=((i-2)*n+j-1)*4+3;
    				T.addEdge(u, nxtv, 0);
    				T.addEdge(nxtv, u, 0);
    			}
    			if(i+1<=n&&map[i+1][j]!='x') {
    				int nxtu=(i*n+j-1)*4+1;
    				T.addEdge(nxtu, v, 0);
    				T.addEdge(v, nxtu, 0);	
    			}	
	    		if(map[i][j]=='A') {
	    			s[0]=l;
	    			s[1]=u;
	    			
	    		}else if(map[i][j]=='B') {
	    			t[0]=l;
	    			t[1]=u;
	    			
	    		}
	    		
	    	}
	    }
	    T.Dij(s[0], N, t);
	    T.Dij(s[1], N, t);
	    if(T.ans==inf)out.print("-1");
	    else out.print(T.ans);
 	    out.flush();
	    out.close();
	}
	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();
	    }
	}
}

UVALive7250 Meeting

题目大意

$n$ 个点， $m$ 个集合，集合内的点有权为 $f$ 的边
$A$ 从 $1$ 出发， $B$ 从 $n$ 出发，走单位距离花费 $1$ 时间，问 $AB$ 相遇与一点的最少时间

解题思路

若点集合 $size=k$ ，任意两点之间连 $f$ 边，共连 $\textrm{C}_{k}^{2}$ ，当 $k$ 很大存不下
考虑将集合看作一点 $t$ ，则集合内两点 $x,y$ 连边，变为 $x\overset{f/2}\leftrightarrow t\overset{f/2}\leftrightarrow y,double \ f$
连边降为 $k*2$ ，可以存下
$A,B$ 在 $n$ 个点上的任意一点相遇
以 $1$ 和 $n$ 为出发点，跑最短路
再枚举点，求 $Min^n_{i=1} \ max(dis_1[i],dis_n[i])$

[USACO 2007 Mar G] Ranking the cows

题目链接

题目大意

有 $n$ 个数字，已经比较了 $m$ 对
问至少还需要多少对比较，可以将 $n$ 个数由大到小排列

解题思路

对于 $x,y,z$ ，若已知 $x>y,y>z$ ，则 $x>z$
$n\leq 1000$ ，类似 $Floyd$ 枚举中间点
定义 $f[i][j]$ 表示 $i>j,1=true,0=false$
$f[i][j]=f[i][k]\&f[k][j],(1\&1=1)$
可以省去枚举 $j$ ， $if(f[i][k])f[i]|=f[k]$
$f[i]$ 表示与其他点的状态，用 $bitset$
若 $f[i][j]==0\&\&f[j][i]==0$ ，则不知道 $i,j$ 大小关系，需要比较， $ans++$
此时不知道比较后的结果，无法传递，要满足至少都得比较


import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.BitSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.Vector;



public class Main{
	
	public static void main(String[] args) throws IOException{
		AReader input=new AReader();
	    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	    int n=input.nextInt();
	    int m=input.nextInt();
	    BitSet[] a=new BitSet[n+1];
	    for(int i=0;i<n;++i)a[i]=new BitSet(n);
	    for(int i=1;i<=m;++i) {
	    	int x=input.nextInt()-1;
	    	int y=input.nextInt()-1;
	    	
	    	a[x].set(y);//对应位置赋为true
	    }
	    for(int k=0;k<n;++k) {
	    	for(int i=0;i<n;++i) {
	    		if(a[i].get(k)) {
	    			a[i].or(a[k]);//或运算
	    		}
	    	}
	    }
	    int ans=0;
	    for(int i=0;i<n;++i) {
	    	for(int j=i+1;j<n;++j) {
	    		if(a[i].get(j)==false&&a[j].get(i)==false) {
	    			ans++;
	    			//因为不知道是i>j还是i<j，无法传递
	    		}
	    	}
	    }
	    out.print(ans);
 	    out.flush();
	    out.close();
	}
	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();
	    }
	}
}

[SCOI2012]滑雪与时间胶囊

题目链接

题目大意

$m$ 条边，每个边的端点为景点，共 $n$ 个
每个点有高度 $H_i$
$i$ 能到 $j$ ，满足有 $i\rightarrow j\&\&H_i\geq H_j$ ，边权为 $w$
从从 $1$ 号景点出发，走最短路，访问尽量多的节点
同时可以回到经过的任意一个节点，再次出发
问最多能访问多少个节点和此时走过的最小路径长度

解题思路

由于可以任意回溯，所以可以访问到 $1$ 号节点能到的任意一个节点
设 $1$ 号节点能到的点数为 $k$ ，则最终边有 $k-1$ 条
考虑 $Prim$ 生成最小树
由于边有高度通行限制，所以要访问尽可能多的点，高度高的点先出，其次再比较边权
每出一个点，进行答案更新

import java.io.*;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.StringTokenizer;




public class Main{
	static int cnt=0;
	static Edge[] e;
	static int[] head;
	static
	class Edge{
		long val;
		int to;
		int fr;
		int nxt;
	}
	static void addEdge(int fr,int to,long val) {
		cnt++;
		e[cnt]=new Edge();
		e[cnt].fr=fr;
		e[cnt].to=to;
		e[cnt].val=val;
		e[cnt].nxt=head[fr];
		head[fr]=cnt;
	}
	static
	class Node{
		int x;
        int h;
		long dis;
		public Node(){}
		public Node(int X,long D,int H){
			x=X;
            h=H;
			dis=D;
		}
	}
	
	
	public static void main(String[] args) throws IOException{
		AReader input=new AReader();
	    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	    int n=input.nextInt();
	    int m=input.nextInt();
	    int[] H=new int[n+1];

	    e=new Edge[m<<1|1];
	    head=new int[n+1];
	    for(int i=1;i<=n;++i) {
	    	H[i]=input.nextInt();

	    }
	    for(int i=1;i<=m;++i) {
	    	int u=input.nextInt();
	    	int v=input.nextInt();
	    	long val=input.nextLong();
	    	if(H[u]>=H[v])addEdge(u, v, val);
	    	if(H[v]>=H[u])addEdge(v, u, val);
	    }
	    boolean[] vis=new boolean[n+1];
	    PriorityQueue<Node> qu=new PriorityQueue<Node>((o1,o2)->{
	    	if(o1.h==o2.h){
                if(o1.dis-o2.dis>0)return 1;
                else if(o1.dis-o2.dis<0)return -1;
                else return 0;
            }else return o2.h-o1.h;
	    });
	    int num=0;
	    long ans=0;
	    qu.add(new Node(1,0,H[1]));
	    while(!qu.isEmpty()) {
	    	Node now=qu.peek();
            qu.poll();
	    	int x=now.x;
            if(vis[x])continue;
	    	vis[x]=true;
	    	ans+=now.dis;
	    	num++;
	    	for(int i=head[x];i>0;i=e[i].nxt) {
	    		int v=e[i].to;
	    		long w=e[i].val;
	    		if(vis[v])continue;
	    		
                qu.add(new Node(v,w,H[v]));

	    	}
	    }
	    out.println(num+" "+ans);
	    out.flush();
	    out.close();
	}
	
	static
    class AReader {
        private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        private StringTokenizer tokenizer = new StringTokenizer("");
        private String innerNextLine() {
            try {
                return reader.readLine();
            } catch (IOException ex) {
                return null;
            }
        }
        public boolean hasNext() {
            while (!tokenizer.hasMoreTokens()) {
                String nextLine = innerNextLine();
                if (nextLine == null) {
                    return false;
                }
                tokenizer = new StringTokenizer(nextLine);
            }
            return true;
        }
        public String nextLine() {
            tokenizer = new StringTokenizer("");
            return innerNextLine();
        }
        public String next() {
            hasNext();
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
    }
}