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);}}@Overridepublic 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);}}@Overridepublic 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();}@Overridepublic void run() {while (scanner.hasNext()) {g[0][0] = scanner.nextInt();g[0][1] = scanner.nextInt();g[0][2] = scanner.nextInt();//bgcg[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];//bcgans[1] = g[1][0] + g[2][0] + g[0][1] + g[2][1] + g[0][2] + g[1][2];//bgcans[2] = g[1][2] + g[2][2] + g[0][0] + g[2][0] + g[0][1] + g[1][1];//cbgans[3] = g[1][2] + g[2][2] + g[0][1] + g[2][1] + g[0][0] + g[1][0];//cgbans[4] = g[1][1] + g[2][1] + g[0][0] + g[2][0] + g[0][2] + g[1][2];//gbcans[5] = g[1][1] + g[2][1] + g[0][2] + g[2][2] + g[0][0] + g[1][0];//gcbint 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();}}
}
