十四届蓝桥杯省赛Java B组 合并区域

就是将两个矩阵进行拼接，两矩阵可以旋转90 180 270 度。

因为数据比较小，所以这基本上就是一个大的枚举模拟加搜索，直接暴力求解。

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


public class Main{
    static int n;
    static int N = 101;
    static int mod = (int)1e9 + 7;
    static StreamTokenizer stt = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static int[][] f = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
//	static int[][] ff = {{0, -1, 0}, {0, 0, 1}, {0, 1, 0}, {0, 0, -1}, {1, 0, 0}, {-1, 0, 0}};
//	static int[] month = {0, 31,28,31,30,31,30,31,31,30,31,30,31};
	static int[][] o = new int[N][N];
	static int[][] p = new int[N][N];
	static int[][] m = new int[3 * N][3 * N];
	static int maxi;
	
	private static int dfs(int x, int y) {
		int k = 1;
		m[x][y] = 0;
		for(int i = 0; i < 4; i ++) {
			int nx = x + f[i][0], ny = y + f[i][1];
			if(nx < 1 || nx > 3 * n || ny < 1 || ny > 3 * n || m[nx][ny] == 0) continue;
			k += dfs(nx, ny);
		}
		return k;
	}
	private static void draw(int bx, int by, int[][] o2) {
		for(int i = bx, ii = 1; ii <= n; ii ++, i ++) {
			for(int j = by, jj = 1; jj <= n; jj ++, j ++) {
				m[i][j] = o2[ii][jj];
			}
		}
	}
	static void asd(int x, int y) throws IOException {
		// 在地图上画o矩阵
		draw(n + 1, n + 1, o);
		// 在地图上画p矩阵
		draw(x, y, p);
		// 分别枚举o矩阵和p矩阵所有的1进行深搜
		for(int i = 1; i <= n; i ++) {
			for(int j = 1; j <= n; j ++) {
				int nx = i + n, ny = j + n;
				if(m[nx][ny] == 1)
					maxi = Math.max(maxi, dfs(nx, ny));
			}
		}
		for(int i = 1; i <= n; i ++) {
			for(int j = 1; j <= n; j++) {
				int nx = i + x - 1, ny = j + y - 1;
				if(m[nx][ny] == 1) 
					maxi = Math.max(maxi, dfs(nx, ny));
			}
		}
		
	}
	
	private static void sov() throws IOException {
		// 在一个三倍大的地图中，枚举p矩阵的坐上角顶点，默认o在中心的n阶矩阵位置
		for(int i = 1; i <= 2 * n + 1; i ++) {
			 asd(1, i);
			 asd(2 * n + 1, i);
			 asd(i, 1);
			 asd(i, 2 * n + 1);
		}
	}
	static void rotate() {
		int[][] s = new int[N][N];
        for(int i = 1; i <= n; i ++) {
        	for(int j = 1; j <= n; j ++) {
        		s[j][n - i + 1] = o[i][j];
        	}
        }
        for(int i = 1; i <= n; i ++) {
        	for(int j = 1; j <= n; j ++) {
        		o[i][j] = s[i][j];
        	}
        }
	}

	static void sovle() throws Exception {
		n = readInt();
		for(int i = 1; i <= n; i ++) {
			for(int j = 1; j <= n; j ++) {
				o[i][j] = readInt();
			}
		}
		for(int i = 1; i <= n; i ++) {
			for(int j = 1; j <= n; j ++) {
				p[i][j] = readInt();
			}
		}
		
		maxi = 0;
//		rotate();
//		for(int i = 1; i <= n; i ++){
//			for(int j = 1; j <= n; j ++) {
//				bw.write(o[i][j] + " ");
//			}
//			bw.write("\n");
//		}
		for(int i = 0; i < 4; i ++) {
			// 旋转o矩阵
			rotate();
			// 拼接两矩阵求解
			sov();
		}
		
		bw.write(maxi + "\n");
	}
	
	
	public static void main(String args[]) throws Exception {
		int t = 1;
//		t = Integer.parseInt(br.readLine());
//		t = readInt();
		while((t --) > 0) {
//		while((n = Integer.parseInt(br.readLine())) != 0) {
			sovle();
		}
		bw.flush();
		bw.close();
	}

	static int readInt() {
		try {
			stt.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (int)stt.nval;
	}
}