1.30

1.30

Maze - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 从一个空格走cnt - k个点并标记，然后将没有标记的点设为A即可

import java.io.BufferedReader;
import java.io.IOException;
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 implements Runnable {

    static Scanner cin = new Scanner(System.in);
    static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
    static int n, m, k, t;
    static final int N = (int) (5e2 + 10);
    static int[][] g = new int[N][N];
    static boolean[][] str = new boolean[N][N];
    static int[] dx = {1, 0, -1, 0}; static int[] dy = {0, 1, 0, -1};

    public static void main(String[] args) {
        new Thread(null, new Main(), "", 1 << 29).start();
    }

    public static void dfs(int x, int y) {
        if(k == 0) return;

        for(int i = 0; i < 4; i++) {
            int nowx = x + dx[i];
            int nowy = y + dy[i];
            if(nowx < 1 || nowx > n || nowy < 1 || nowy > m) continue;
            if(str[nowx][nowy] || g[nowx][nowy] == 1 || k == 0) continue;
            str[nowx][nowy] = true; k--;
            dfs(nowx, nowy);
        }
    }


    @Override
    public void run() {
        n = cin.nextInt();m = cin.nextInt();k = cin.nextInt();
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            String s = cin.next();
            for (int j = 1; j <= m; j++) {
                if (s.charAt(j - 1) == '#') {
                    g[i][j] = 1;
                } else {
                    cnt++;
                    g[i][j] = 0;
                }
            }
        }
        k = cnt - k;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (!str[i][j] && g[i][j] == 0 && k != 0) {
                    k--;str[i][j] = true;dfs(i, j);
                }
            }
        }

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (!str[i][j] && g[i][j] == 0) {
                    cout.print('X');continue;
                }
                if (g[i][j] == 0) cout.print('.');
                else cout.print('#');
            }
            cout.println();
        }

        cout.flush();
    }
}

New Reform - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 本题可以用并查集来做，但是因为是练习搜索，姑且还是先用搜索来做
• 1 考虑图为一个连通块，图的形式其实就只有三种形态，可以看到分别是不是环，一个环，多个环。可以看到，只有不是环的时候，才会出现一个入度为0的点，也就是根节点。

• 2 考虑图为多个连通块，那么会发现答案为2，就是无环的两个根节点

• 那么ok，答案已经出来了，就是遍历所有结点，对没有被标记过的结点进行递归，寻找这个连通块是否有环即可，注意是单向联通

package shuati;

import java.io.BufferedReader;
import java.io.IOException;
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 implements Runnable {

    static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
    static int n, m;
    static final int N = (int) (1e5 + 10);
    static Vector<Integer>[] city = new Vector[N];
    static boolean[] str = new boolean[N];
    static boolean isCircle;

    public static void main(String[] args) {
        new Thread(null, new Main(), "", 1 << 29).start();
    }

    public static int nextInt() throws IOException {
        cin.nextToken();
        return (int) cin.nval;
    }

    public static void dfs(int son, int parent) {
        for (Integer now : city[son]) {
            if (now == parent) continue;
            if (str[now]) {
                isCircle = true;
                continue;
            }
            str[now] = true;
            dfs(now, son);
        }
    }


    @Override
    public void run() {
        try {
            for(int i = 0; i < N; i++) {city[i] = new Vector<>();}
            n = nextInt();m = nextInt();
            for(int i = 1; i <= m; i++) {int x = nextInt();int y = nextInt();city[x].add(y);city[y].add(x);}
            int ans = 0;
            for(int i = 1; i <= n; i++) {
                if(!str[i]) {
                    str[i] = true;
                    isCircle = false;
                    dfs(i, -1);
                    if(!isCircle) ans++;
                }
            }
            cout.println(ans);cout.flush();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }

    }
}

Ecological Bin Packing - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 数据比较简单，大模拟即可

import java.io.BufferedReader;
import java.io.IOException;
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 implements Runnable {

    static Scanner scanner = new Scanner(System.in);
    static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
    static int[][] g = new int[3][3];
    static String[] s = {"BCG", "BGC", "CBG", "CGB", "GBC", "GCB"};
    static int[] ans = new int[6];


    public static void main(String[] args) {
        new Thread(null, new Main(), "", 1 << 29).start();
    }


    @Override
    public void run() {
        while (scanner.hasNext()) {
            g[0][0] = scanner.nextInt();g[0][1] = scanner.nextInt();g[0][2] = scanner.nextInt();//bgc
            g[1][0] = scanner.nextInt();g[1][1] = scanner.nextInt();g[1][2] = scanner.nextInt();
            g[2][0] = scanner.nextInt();g[2][1] = scanner.nextInt();g[2][2] = scanner.nextInt();
            ans[0] = g[1][0] + g[2][0] + g[0][2] + g[2][2] + g[0][1] + g[1][1];//bcg
            ans[1] = g[1][0] + g[2][0] + g[0][1] + g[2][1] + g[0][2] + g[1][2];//bgc
            ans[2] = g[1][2] + g[2][2] + g[0][0] + g[2][0] + g[0][1] + g[1][1];//cbg
            ans[3] = g[1][2] + g[2][2] + g[0][1] + g[2][1] + g[0][0] + g[1][0];//cgb
            ans[4] = g[1][1] + g[2][1] + g[0][0] + g[2][0] + g[0][2] + g[1][2];//gbc
            ans[5] = g[1][1] + g[2][1] + g[0][2] + g[2][2] + g[0][0] + g[1][0];//gcb
            int k = ans[0];int t = 0;
            for(int i = 1; i <= 5; i++) {
                if(k > ans[i]) {
                    k = ans[i];t = i;
                }
            }

            cout.println(s[t] + " " + k);
            cout.flush();
        }

    }
}