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P1162 填涂颜色 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 只需要环最外圈的0，然后标记，最后填色时没有标记的标为2即可

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;

public class Main {

	static final int N = 35;
	static int[][] g = new int[N][N];
	static boolean[][] str = new boolean[N][N];
	static int n;
	static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
	static int[] dx = {1, 0, -1, 0};
	static int[] dy = {0, 1, 0, -1};
	static Queue<Integer> queue = new LinkedList<>();
	
	public static void dfs(int x, int y) {
		for(int i = 0; i < 4; i++) {
			int nowx = dx[i] + x;
			int nowy = dy[i] + y;
			if(nowx >= 1 && nowx <= n && nowy >= 1 && nowy <= n) {
				if(g[nowx][nowy] == 0 && str[nowx][nowy] == false) {
					str[nowx][nowy] = true;
					dfs(nowx, nowy);
				}
			}
		}
	}

	public static void bfs(int x, int y) {

	}
	
	public static void main(String[] args) throws IOException{
		cin.nextToken();
		n = (int)cin.nval;
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= n; j++) {
				cin.nextToken();
				g[i][j] = (int)cin.nval;
			}
		}
		
		for(int i = 1; i <= n; i++) {
			if(g[1][i] == 0 && str[1][i] == false) {
				str[1][i] = true;
				dfs(1, i);
			}
		}
		
		for(int i = 1; i <= n; i++) {
			if(g[n][i] == 0 && str[n][i] == false) {
				str[n][i] = true;
				dfs(n, i);
			}
		}
		
		for(int i = 2; i <= n - 1; i++) {
			if(g[i][1] == 0 && str[i][1] == false) {
				str[i][1] = true;
				dfs(i, 1);
			}
			
			if(g[i][n] == 0 && str[i][n] == false) {
				str[i][n] = true;
				dfs(i, n);
			}
		}
		
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= n; j++) {
				if(g[i][j] == 0 && str[i][j] == false) {
					g[i][j] = 2;
				}
				System.out.print(g[i][j] + " ");
			}
			System.out.println();
		}
	}
}

[P1460 USACO2.1] 健康的荷斯坦奶牛 Healthy Holsteins - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 注意几个剪枝的点，第一个是大于或等于最短的成功路径不必继续搜，主循环搜完的点不能再搜，因为我们是从1开始的，所以每一次我们都默认往后搜而不往前搜，往前搜必然是与之前的成功路径重复了，再剪一次就能够ac了。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;

public class Main {


	static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
	static int v, g;//4 3
	static final int N = (int)1e3;
	static int[] wtmin_target = new int[N];
	static int[] wtmin_now = new int[N];
	static int[][] siliao = new int[N][N];
	static boolean[] str = new boolean[N];
	static PriorityQueue<Integer>[] targetQueue = new PriorityQueue[N];
	static PriorityQueue<Integer> nowQueue = new PriorityQueue();
	static List<Integer> lists = new ArrayList<>();
	static int cnt = 0x3f3f;
	static int activeMent;

	public static void dfs(int x, int cnt_) {
		if(cnt <= cnt_) {
			return;
		}
		int num = 0;
		for(int i = 1; i <= v; i++) {

			if(wtmin_now[i] >= wtmin_target[i]) {
				num++;
			}
			if(num == v) {
				for (Integer integer : nowQueue) {
					targetQueue[activeMent].add(integer);
				}
				activeMent++;
				cnt = Math.min(cnt, cnt_);
				return;
			}
		}

		for(int i = x + 1; i <= g; i++) {
			if(str[i] == false) {
				str[i] = true;
				for(int j = 1; j <= v; j++) {
					wtmin_now[j] += siliao[i][j];
				}
				nowQueue.add(i);
				dfs(i, cnt_ + 1);
				nowQueue.remove(i);
				for(int j = 1; j <= v; j++) {
					wtmin_now[j] -= siliao[i][j];
				}
				str[i] = false;
			}
		}
	}

	public static void bfs(int x, int y) {

	}
	
	public static void main(String[] args) throws IOException {
		for (int i = 0; i < N; i++) {
			targetQueue[i] = new PriorityQueue<>();
		}

		cin.nextToken();
		v = (int)cin.nval;
		for (int i = 1; i <= v; i++) {
			cin.nextToken();
			wtmin_target[i] = (int) cin.nval;
		}
		cin.nextToken();
		g = (int) cin.nval;

		for (int i = 1; i <= g; i++) {
			for (int j = 1; j <= v; j++) {
				cin.nextToken();
				siliao[i][j] = (int) cin.nval;
			}
		}

		for(int i = 1; i <= g; i++) {
			str[i] = true;
			for(int j = 1; j <= v; j++) {
				wtmin_now[j] += siliao[i][j];
			}
			nowQueue.add(i);
			dfs(i, 1);
			nowQueue.remove(i);
			for(int j = 1; j <= v; j++) {
				wtmin_now[j] -= siliao[i][j];
			}
			//str[i] = false;
		}

		int xiabiao = 0;
		for(int i = 0; i < activeMent; i++) {
			if (targetQueue[i].size() == cnt) {
				xiabiao = i;
				break;
			}
		}

		System.out.print(cnt + " ");

		while(targetQueue[xiabiao].size() != 0) {
			System.out.print(targetQueue[xiabiao].poll() + " ");
		}


	}
}

[P1030 NOIP2001 普及组] 求先序排列 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 每一次递归输出后序遍历最后一个节点，然后将中序分为两个子串，然后按照后续顺序，分为两个后序子串，因为个数肯定一样，可以使用java的substr，写者有点rz的没有想到二者两个字母个数是完全一样的，调了一个小时。然后分别递归dfs即可。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;

public class Main {


	static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
	static String zhongxu = new String();
	static String houxu = new String();

	// s中序 c后序
	public static void dfs(String s, String c) {
		if(s.length() == 1) {
			System.out.print(c.charAt(c.length() - 1));
			return;
		}
		System.out.print(c.charAt(c.length() - 1));
		String[] t = s.split(Character.toString(c.charAt(c.length() - 1))); //t0 t1 中序

		String houxu1 = new String();
		String houxu2 = new String();
		for(int i = 0; i < c.length() - 1; i++) {
			if(t[0].contains(Character.toString(c.charAt(i)))) {
				houxu1 += Character.toString(c.charAt(i));
			} else {
				houxu2 += Character.toString(c.charAt(i));
			}
		}


		if(t.length == 2 && t[0].length() != 0 && t[1].length() != 0) {
			dfs(t[0], houxu1);
			dfs(t[1], houxu2);
		} else {
			if(houxu1.length() != 0) {
				dfs(t[0], houxu1);
			} else {
				dfs(t[1], houxu2);
			}
		}

	}

	public static void main(String[] args) throws Exception{
		cin.nextToken();
		zhongxu = cin.sval;
		cin.nextToken();
		houxu = cin.sval;


		System.out.print(houxu.charAt(houxu.length() - 1));
		String[] t = zhongxu.split(Character.toString(houxu.charAt(houxu.length() - 1)));

		String houxu1 = new String();
		String houxu2 = new String();
		for(int i = 0; i < houxu.length() - 1; i++) {
			if(t[0].contains(Character.toString(houxu.charAt(i)))) {
				houxu1 += Character.toString(houxu.charAt(i));
			} else {
				houxu2 += Character.toString(houxu.charAt(i));
			}
		}

		if(t.length == 2 && t[0].length() != 0 && t[1].length() != 0) {
			dfs(t[0], houxu1);
			dfs(t[1], houxu2);
		} else {
			if(houxu1.length() != 0) {
				dfs(t[0], houxu1);
			} else {
				dfs(t[1], houxu2);
			}
		}
	}
}