hn 10432 棋盘覆盖

灰太狼 2022-01-07 13:13 185阅读 0赞

<table> 
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 <td>棋盘覆盖问题</td> 
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 <td>Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB</td> 
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 <td>Total submit users: 65, Accepted users: 26</td> 
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 <td>Problem 10432 : No special judgement</td> 
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 <td>Problem description</td> 
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 <td>&nbsp;&nbsp;在一个2k x 2k ( 即：2^k x 2^k )个方格组成的棋盘中，恰有一个方格与其他方格不同，称该方格为一特殊方格，且称该棋盘为一特殊棋盘。在棋盘覆盖问题中，要用图示的4种不同形态的L型骨牌覆盖给定的特殊棋盘上除特殊方格以外的所有方格，且任何2个L型骨牌不得重叠覆盖。 <img src="http://acm.hunnu.edu.cn/online/images/10432.jpg" alt="" width="384" height="101"> </td> 
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 <td>Input</td> 
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 <td>&nbsp;&nbsp;输入文件第一行是一个整数T,表示有多少组测试数据，接下来是T组测试数据，共2T行，每组第一行为整数n,是2的n次幂（1&lt;=n&lt;=64），表示棋盘的大小为n*n,第二行是两个整数，代表特殊方格所在行号和列号。 </td> 
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 <td>Output</td> 
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 <td>&nbsp;&nbsp;先输出“CASE:i，然后按样例输出。数据间用制表符隔开(‘t’),每行最后一个数据后无制表符。 </td> 
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 <td>Sample Input</td> 
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 <td> <pre>2
2
0 0
8
2 2
</pre> </td> 
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 <td>Sample Output</td> 
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 <td> <pre>CASE:1
0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1
1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1
CASE:2
3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 8&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 8&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 9&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 9
3 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 8&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 7&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 7&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 9
5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 7&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 11
5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 11&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 11
13&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 13&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 14&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 18&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 19&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 19
13&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 12&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 14&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;14&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 18&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 18&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 17&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 19
15&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 12&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 12&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 16&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 20&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 17&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 17&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 21
15&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 15&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 16&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 16&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 20&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 20&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 21&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 21
</pre> </td> 
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 <td>Judge Tips</td> 
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 <td> &nbsp;&nbsp;要求遍历顺序按从左到右，从上到下。 &nbsp; 
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 <pre>package acm.icpc;

import java.util.Scanner;
/**
 * @功能Function Description: 棋盘覆盖
 * @开发环境Environment: eclipse
 * @技术特点Technique: 
 * @算法 
 * @版本Version: Scanner
 * @作者Author: follow your dreams
 * @日期Date: 20120809
 * @备注Notes: 测试数据都对了，可是提交一直wrong error
 * @链接： http://acm.hunnu.edu.cn/online/?action=problem&amp;type=show
 */
public class Icpc10432_1_20120820 {
 
 public static int[][] arr;
 private static int num = 1;//静态变量
 public static void main(String[] args) {
 Scanner sc = new Scanner(System.in);
 int count = sc.nextInt();
 int countCase = 0;
 while(count-- != 0) {
 countCase ++;
 int n = sc.nextInt(); 
 int x = sc.nextInt();
 int y = sc.nextInt();
 arr = new int[n+1][n+1];
 num = 1;
 if(n==1) {
 System.out.println("0");
 } else {
 solve(1, 1, x+1, y+1, n);
 show(n, countCase);
 }
 }
 }
 
 /**
 * 将棋盘分割成4个部分
 * @param i 棋盘的左上角的坐标(i,j)
 * @param j 
 * @param x 特殊的那个位置
 * @param y
 * @param n 棋盘的宽度（由于棋盘是正方形的，有宽度和左上角的坐标就可以确定棋盘了）
 */
 public static void solve(int i, int j, int x, int y, int n) {
 int temp = n/2;
 int flag = 0;
 int xtemp1 = i+temp-1;
 int ytemp1 = j+temp-1;
 if(!(x&lt;=xtemp1 &amp;&amp; y&lt;=ytemp1)) {
 arr[i+temp-1][j+temp-1] = num;
 } else {
 flag = 1;
 }
 int xtemp2 = i+temp-1;
 int ytemp2 = j+temp;
 if(!(x&lt;=xtemp2 &amp;&amp; y&gt;=ytemp2)) {
 arr[xtemp2][ytemp2] = num;
 } else {
 flag = 2;
 }
 int xtemp3 = i+temp;
 int ytemp3 = j+temp-1;
 if(!(x&gt;=xtemp3 &amp;&amp; y&lt;=ytemp3)) {
 arr[xtemp3][ytemp3] = num;
 } else {
 flag = 3;
 }
 int xtemp4 = i+temp;
 int ytemp4 = j+temp;
 if(!(x&gt;=xtemp4 &amp;&amp; y&gt;=ytemp4)) {
 arr[xtemp4][ytemp4] = num;
 } else {
 flag = 4;
 }
 num ++; 
 if(n/2 != 1) {
 if(flag != 1) {
 solve(i, j, xtemp1, ytemp1, temp);
 } else {
 solve(i, j, x, y, temp);
 }
 if(flag != 2) {
 solve(i, temp+j, xtemp2, ytemp2, temp);
 } else {
 solve(i, temp+j, x, y, temp);
 }
 if(flag != 3) {
 solve(temp+i, j, xtemp3, ytemp3, temp);
 } else {
 solve(temp+i, j, x, y, temp);
 }
 if(flag != 4) {
 solve(temp+i, temp+j, xtemp4, ytemp4, temp);
 } else {
 solve(temp+i, temp+j, x, y, temp);
 }
 }
 }
 
 public static void show(int n, int countCase) {
 //输出
 int i, j;
 System.out.println("CASE:" + countCase);
 for(i=1; i&lt;=n; i++) {
 for(j=1; j&lt;n; j++) {
 System.out.printf("%d\t", arr[i][j]);
 }
 System.out.println(arr[i][j]);
 }
 }

}</pre> 
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转载于:https://www.cnblogs.com/followYourDreams/archive/2012/08/21/2648569.html