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做三个月网站广告收入企业如何进行网络推广

news 2026/4/7 10:40:41

做三个月网站广告收入,企业如何进行网络推广,合肥专门做网站的公司有哪些,博罗网站建设Going Home Here 解题思路模板二分图最优匹配，前提是有完美匹配（即存在一一配对）左右集合分别有顶标，当时，为有效边，即选中初始对于左集合每个点，选择其连边中最优的，然后对于每…

Going Home

Here

解题思路

$KM$ 模板
二分图最优匹配，前提是有完美匹配（即存在一一配对）
左右集合分别有顶标 $lx[i],ly[j]$ ，当 $lx[i]+ly[j]=w[i][j]$ 时， $i\rightarrow j$ 为有效边，即选中
初始 $lx[i]=-inf,ly[j]=0$
对于左集合每个点，选择其连边中最优的， $lx[i]=Max\ w$
然后对于每个点找其最优匹配
若能在前面连好边的图中找到匹配，则继续下一个点
若不能，考虑撤销边
如何撤销？
对于这次找增广路左集合中的点，找一条不在这条增广路上的边，进行替换
找到最小代价，即 $d=Min\ lx[i]+ly[j]-w[i][j]$
对于这次找增广路访问到的左集合 $lx[i]-d$ ，右集合 $ly[j]+d$
重复进行，直到实现该点找到增广路可以添入或不能进行替换（无完美匹配）
对于该题，求最小，只用将距离变为负值，最后结果在反回来
每次找增广路前清空 $visx[],viy[]$


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;//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=200;staticclass Node{int x;int y;public Node(int u,int v) {x=u;y=v;}}static int[][] w=new int[N+1][N+1];static void getw(Node a,Node b,int i,int j) {w[i][j]=-(Math.abs(a.x-b.x)+Math.abs(a.y-b.y));}static int mm=0;static int hh=0;static int[] lx=new int[N+1];static int[] ly=new int[N+1];static int[] link=new int[N+1];static boolean[] visx=new boolean[N+1];static boolean[] visy=new boolean[N+1];static Node[] men=new Node[N+1];static Node[] home=new Node[N+1];static boolean dfs(int x) {visx[x]=true;for(int i=1;i<=hh;++i) {if(!visy[i]&&lx[x]+ly[i]==w[x][i]) {visy[i]=true;if(link[i]==0||dfs(link[i])) {link[i]=x;return true;}}}return false;}static void solve() throws Exception{AReader input=new AReader();PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));	while(true) {int n=input.nextInt();int m=input.nextInt();if(n==0&&m==0)break;mm=0;hh=0;Arrays.fill(link, 0);for(int i=1;i<=n;++i) {String string=" "+input.next();char[] mp=string.toCharArray();for(int j=1;j<mp.length;++j) {if(mp[j]=='m') {mm++;men[mm]=new Node(i, j);}else if(mp[j]=='H') {hh++;home[hh]=new Node(i, j);}}}for(int i=1;i<=mm;++i){for(int j=1;j<=hh;++j) {getw(men[i], home[j], i, j);}}for(int i=1;i<=mm;++i) {lx[i]=-inf;for(int j=1;j<=hh;++j) {ly[j]=0;lx[i]=Math.max(lx[i], w[i][j]);}}boolean ok=true;for(int i=1;i<=mm;++i) {while(true) {//对于当前点i找个匹配,一直试Arrays.fill(visx, false);Arrays.fill(visy, false);int dis=inf;if(dfs(i))break;//直接不能找到//考虑改边for(int j=1;j<=mm;++j) {if(!visx[j]) continue;for(int k=1;k<=hh;++k) {if(visy[k])continue;dis=Math.min(dis, (lx[j]+ly[k])-w[j][k]);}}if(dis==inf) {ok=false;break;}for(int j=1;j<=mm;++j)if(visx[j])lx[j]-=dis;for(int j=1;j<=hh;++j)if(visy[j])ly[j]+=dis;}if(!ok)break;}int sum=0;if(ok) {for(int i=1;i<=hh;++i) {sum-=w[link[i]][i];}out.println(sum);}else out.println(-1);}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) {
//			}
//		}staticclass 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();}}}

poj3041 Asteroids

Here

解题思路

对于一个陨石 $(i,j)$ ，可以选择打第 $i$ 行或第 $j$ 列，一定只选其中一个
考虑二分图匹配
左集合为所有行，右集合为所有列，每个陨石看作行列的一条连边
所以打完陨石的最少次数，即选择最少的点实现对所有边覆盖
即求最小点覆盖，也就是最大匹配（最小点覆盖等于最大匹配）


import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
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=501;static int n=0;static int k=0;static boolean[][] w=new boolean[N+1][N+1];static boolean[] visy=new boolean[N+1];static int[] link=new int[N+1];static boolean dfs(int x) {for(int i=1;i<=n;++i) {if(!visy[i]&&w[x][i]) {visy[i]=true;if(link[i]==0||dfs(link[i])) {link[i]=x;return true;}}}return false;}static void solve() throws Exception{AReader input=new AReader();PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));	n=input.nextInt();k=input.nextInt();for(int i=1;i<=k;++i) {int x=input.nextInt();int y=input.nextInt();w[x][y]=true;}int sum=0;for(int i=1;i<=n;++i) {Arrays.fill(visy, false);if(dfs(i))sum++;}out.print(sum);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) {
//			}
//		}staticclass 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();}}}

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