1.22

P1123 取数游戏 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 本题一开始是想写双for循环，复杂度\(O(n^n)\)，不出意外T了三个点，然后思考剪枝，考虑到不需要重新找，做了一个bool数组来标记，防止重新搜索已经出现过的方案，过了第四个点。

• 后面看了题解后，发现这一思路极好，枚举整个棋盘，每个点可以选也可以不选，则复杂度直接降为\(O(2^n)\),直接ac，根本不用剪枝

package shuati;

import java.io.*;
import java.util.*;

public class Main {

    static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static Scanner scanner = new Scanner(System.in);
    static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
    static int t, n, m;
    static final int N = 10;
    static int[][] g = new int[N][N];
    // cnt[i][j] 表示以（i，j）为中心的周围八个格子有几个数被选取
    static int[][] cnt = new int[N][N];
    static int ans, nowans;
    static int dx[] = {0, 1, 1, 0, -1, -1, -1, 0, 1};
    static int dy[] = {0, 0, 1, 1, 1, 0, -1, -1, -1};

    public static void dfs(int x, int y) {
       // 处理越界
       if(y == m + 1) {
          dfs(x + 1, 1);
          return;
       }
       if(x == n + 1) {
          ans = Math.max(nowans, ans);
          return;
       }
       // 不选择这个点
       dfs(x,y + 1);
       if(cnt[x][y] == 0) {
          for(int i = 0; i < 9; i++) {
             cnt[x + dx[i]][y + dy[i]]++;
          }
          nowans += g[x][y];
          dfs(x, y + 1);
          nowans -= g[x][y];
          for(int i = 0; i < 9; i++) {
             cnt[x + dx[i]][y + dy[i]]--;
          }
       }

    }

    public static void solve() throws IOException {
       for (int i = 1; i <= n; i++) {
          for(int j = 1; j <= m; j++) {
             cin.nextToken();
             g[i][j] = (int) cin.nval;
            }
       }

       dfs(1, 1);
       cout.println(ans);
       cout.flush();
    }

    public static void main(String[] args) throws IOException{
       cin.nextToken();
       t = (int) cin.nval;
       while(t-- != 0) {
          cin.nextToken();
          n = (int) cin.nval;
          cin.nextToken();
          m = (int) cin.nval;
          ans = 0;
          nowans = 0;
          solve();
       }
    }
}

深さ優先探索 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 经典搜索

import java.io.*;
import java.util.*;

public class Main {

	static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static Scanner scanner = new Scanner(System.in);
	static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
	static final int N = 510;
	static int n, m;//长和宽
	static char[][] g = new char[N][N];
	static boolean[][] str = new boolean[N][N];
	static int[] dx = {1, 0, -1, 0};
	static int[] dy = {0, 1, 0, -1};
	static int cnt;

	public static void dfs(int x, int y) {
		if(g[x][y] == 'g') {
			cnt = 1;
			cout.print("Yes");
			return;
		}

		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 <= m) {
				if(str[nowx][nowy] == false && g[nowx][nowy] != '#' && cnt == 0) {
					str[nowx][nowy] = true;
					dfs(nowx, nowy);
				}
			}
		}
	}

	public static void main(String[] args) throws IOException{
//		cin.nextToken();
//		n = (int) cin.nval;
//		cin.nextToken();
//		m = (int) cin.nval;
		n = scanner.nextInt();
		m = scanner.nextInt();
		int chux = 0;
		int chuy = 0;
		for(int i = 1; i <= n; i++) {
			String s = scanner.next();
			for(int j = 1; j <= m; j++) {
				g[i][j] = s.charAt(j - 1);
				if(s.charAt(j - 1) == 's') {
					chux = i;
					chuy = j;
				}
			}
		}

		str[chux][chuy] = true;
		dfs(chux, chuy);
		if(cnt == 0) {
			cout.print("No");
		}
		cout.flush();

    }
}

P1706 全排列问题 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 水题

import java.io.*;
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 n;
	static boolean[] str = new boolean[10];
	static Queue<Integer> myQueue = new LinkedList<>();

	public static void dfs(int cnt) {
		if(cnt == n) {
			for (Integer i : myQueue) {
				cout.print("    " + i);
			}
			cout.println();
			cout.flush();
			return;
		}

		for(int i = 1; i <= n; i++) {
			if(str[i] == false) {
				str[i] = true;
				cnt++;
				myQueue.add(i);
				dfs(cnt);
				myQueue.remove(i);
				cnt--;
				str[i] = false;
			}
		}
	}

	public static void main(String[] args) throws IOException{
		cin.nextToken();
		n = (int)cin.nval;

		dfs(0);
	}
}