蓝桥杯 第七届省赛试题 剪邮票

<div id="post_content_86165038147" class="d_post_content j_d_post_content  clearfix">            剪邮票如【图1.jpg】, 有12张连在一起的12生肖的邮票。现在你要从中剪下5张来，要求必须是连着的。（仅仅连接一个角不算相连）比如，【图2.jpg】，【图3.jpg】中，粉红色所示部分就是合格的剪取。请你计算，一共有多少种不同的剪取方法。请填写表示方案数目的整数。注意：你提交的应该是一个整数，不要填写任何多余的内容或说明性文字。<img style="cursor: url("http://tb2.bdstatic.com/tb/static-pb/img/cur_zin.cur"), pointer;" class="BDE_Image" src="http://imgsrc.baidu.com/forum/w%3D580/sign=2e6a1bf805087bf47dec57e1c2d2575e/f7a2e8c451da81cb0a24a77e5566d0160b24318a.jpg" height="319" width="412" alt="" /><img style="cursor: url("http://tb2.bdstatic.com/tb/static-pb/img/cur_zin.cur"), pointer;" class="BDE_Image" src="http://imgsrc.baidu.com/forum/w%3D580/sign=26cc5ec927a446237ecaa56aa8237246/293cb7025aafa40f0015ea6eac64034f7af019f9.jpg" height="319" width="412" alt="" /><img style="cursor: url("http://tb2.bdstatic.com/tb/static-pb/img/cur_zin.cur"), pointer;" class="BDE_Image" src="http://imgsrc.baidu.com/forum/w%3D580/sign=de82bcb75e82b2b7a79f39cc01accb0a/58c44ef40ad162d98d1546af16dfa9ec8813cdd2.jpg" height="319" width="412" alt="" /></div>package com.diqijie.shengsai;import java.util.Arrays;import java.util.HashSet;import java.util.Iterator;/** * @author leibaobao 剪邮票  * 解： 深搜，效率不是很高，反正是填空题 */public class _7 {public static int count = 0;public static int [] a = new int[5];public static HashSet<String> hashset = new HashSet<String>();public static void main(String[] args) {// TODO Auto-generated method stubfor(a[0] = 0; a[0] < 12; a[0]++)for(a[1] = a[0]+1; a[1] < 12; a[1]++)for(a[2] = a[1]+1; a[2] < 12; a[2]++)for(a[3] = a[2]+1; a[3] < 12; a[3]++)for(a[4] = a[3]+1; a[4] < 12; a[4]++)if(jus()){hashset.add(""+a[0]+a[1]+a[2]+a[3]+a[4]);}System.out.println(hashset.size());}private static boolean jus() {// TODO Auto-generated method stubboolean flag[] = new boolean[5];dfs(flag,0);return flag[0]&&flag[1]&&flag[2]&&flag[3]&&flag[4];}private static void dfs(boolean[] flag,int n) {// TODO Auto-generated method stubflag[n] = true;for(int i = 0; i < 5; i++){//加一减一要在同一行、加四减四要在同一列if(!flag[i] && (a[i]/4 == a[n]/4) && (a[i] == a[n] - 1 || a[i] == a[n] + 1)){dfs(flag,i);}if (!flag[i] && (a[i]%4 == a[n]%4) && (a[i] == a[n] - 4 || a[i] == a[n] + 4)){dfs(flag,i);}}}}