[暴力搜索] POJ 3087 Shuffle'm Up
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10003 | Accepted: 4631 |
Description
A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuffling chips is performed by starting with two stacks of poker chips, S1 and S2, each stack containing C chips. Each stack may contain chips of several different colors.
The actual shuffle operation is performed by interleaving a chip from S1 with a chip from S2 as shown below for C = 5:

The single resultant stack, S12, contains 2 * C chips. The bottommost chip of S12 is the bottommost chip from S2. On top of that chip, is the bottommost chip from S1. The interleaving process continues taking the 2nd chip from the bottom of S2 and placing that on S12, followed by the 2nd chip from the bottom of S1 and so on until the topmost chip from S1 is placed on top of S12.
After the shuffle operation, S12 is split into 2 new stacks by taking the bottommost C chips from S12 to form a new S1 and the topmost C chips from S12 to form a new S2. The shuffle operation may then be repeated to form a new S12.
For this problem, you will write a program to determine if a particular resultant stack S12 can be formed by shuffling two stacks some number of times.
Input
The first line of input contains a single integer N, (1 ≤ N ≤ 1000) which is the number of datasets that follow.
Each dataset consists of four lines of input. The first line of a dataset specifies an integer C, (1 ≤ C ≤ 100) which is the number of chips in each initial stack (S1 and S2). The second line of each dataset specifies the colors of each of the C chips in stack S1, starting with the bottommost chip. The third line of each dataset specifies the colors of each of the C chips in stack S2 starting with the bottommost chip. Colors are expressed as a single uppercase letter (A through H). There are no blanks or separators between the chip colors. The fourth line of each dataset contains 2 * C uppercase letters (A through H), representing the colors of the desired result of the shuffling of S1 and S2 zero or more times. The bottommost chip’s color is specified first.
Output
Output for each dataset consists of a single line that displays the dataset number (1 though N), a space, and an integer value which is the minimum number of shuffle operations required to get the desired resultant stack. If the desired result can not be reached using the input for the dataset, display the value negative 1 (−1) for the number of shuffle operations.
Sample Input
2
4
AHAH
HAHA
HHAAAAHH
3
CDE
CDE
EEDDCC
Sample Output
1 2
2 -1
Source
#include<stdio.h>
#include<string.h>
int main()
{
int T,n,times,i,j,cnt,flag;
char s1[210],s2[210],s12[210],aim[210],temp[210];
scanf("%d",&T);
for(j=1;j<=T;++j)
{
cnt=flag=0;times=1;
scanf("%d%s%s%s",&n,s1,s2,aim);
for(i=0;i<strlen(s2);++i) s12[i*2]=s2[i];
for(i=0;i<strlen(s1);++i) s12[i*2+1]=s1[i];
s12[2*n]='\0';strcpy(temp,s12);
while(strcmp(s12,aim)!=0)
{
strncpy(s1,s12,n);strncpy(s2,s12+n,n);
for(i=0;i<strlen(s2);++i) s12[i*2]=s2[i];
for(i=0;i<strlen(s1);++i) s12[i*2+1]=s1[i];
s12[2*n]='\0';
if(!strcmp(temp,s12))
{
flag=1;
break;
}
++times;
}
if(!flag) printf("%d %d\n",j,times); else printf("%d -1\n",j);
}
return 0;
}
[暴力搜索] POJ 3087 Shuffle'm Up的更多相关文章
- POJ 3087 Shuffle'm Up(洗牌)
POJ 3087 Shuffle'm Up(洗牌) Time Limit: 1000MS Memory Limit: 65536K Description - 题目描述 A common pas ...
- DFS POJ 3087 Shuffle'm Up
题目传送门 /* 题意:两块扑克牌按照顺序叠起来后,把下半部分给第一块,上半部给第二块,一直持续下去,直到叠成指定的样子 DFS:直接模拟搜索,用map记录该字符串是否被搜过.读懂题目是关键. */ ...
- POJ.3087 Shuffle'm Up (模拟)
POJ.3087 Shuffle'm Up (模拟) 题意分析 给定两个长度为len的字符串s1和s2, 接着给出一个长度为len*2的字符串s12. 将字符串s1和s2通过一定的变换变成s12,找到 ...
- POJ 3087 Shuffle'm Up
Shuffle'm Up Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- POJ 3087 Shuffle'm Up 线性同余,暴力 难度:2
http://poj.org/problem?id=3087 设:s1={A1,A2,A3,...Ac} s2={Ac+1,Ac+2,Ac+3,....A2c} 则 合在一起成为 Ac+1,A1,Ac ...
- POJ 3087 Shuffle'm Up 模拟,看着不像搜索啊
题意:给定s1,s1两副扑克,顺序从下到上.依次将s2,s1的扑克一张一张混合.例如s1,ABC; s2,DEF. 则第一次混合后为DAEBFC. 然后令前半段为s1, 后半段为s2. 如果可以变换成 ...
- poj 3087 Shuffle'm Up ( map 模拟 )
题目:http://poj.org/problem?id=3087 题意:已知两堆牌s1和s2的初始状态, 其牌数均为c,按给定规则能将他们相互交叉组合成一堆牌s12,再将s12的最底下的c块牌归为s ...
- POJ 3087 Shuffle'm Up (模拟+map)
题目链接:http://poj.org/problem?id=3087 题目大意:已知两堆牌s1和s2的初始状态, 其牌数均为c,按给定规则能将他们相互交叉组合成一堆牌s12,再将s12的最底下的c块 ...
- POJ 3087 Shuffle'm Up DFS
link:http://poj.org/problem?id=3087 题意:给你两串字串(必定偶数长),按照扑克牌那样的洗法(每次从S2堆底中拿第一张,再从S1堆底拿一张放在上面),洗好后的一堆可以 ...
随机推荐
- Trading
http://v.youku.com/v_show/id_XMTA0OTcxMjgw.html?from=y1.2-1-87.3.8-1.1-1-1-7
- 轻松解决U盘中病毒,文件变成.exe执行文件的问题
U盘中的文件都变成.exe可执行文件是怎么回事?告诉你,你的U盘中病毒了,那么如何清除呢?小编现在就告诉你几个简单方法,轻松就能搞定U盘中病毒问题. 方法1: (1)首先使用杀毒软件把U盘杀杀毒,除去 ...
- Java开发中经典的小实例-(打印输入重复的值)
import java.util.ArrayList;import java.util.Scanner;public class Test8 { public static void main( ...
- .Net架构必备工具列表
★微软MSDN:每个开发人员现在应该下载的十种必备工具 点此进入 ★网友总结.Net架构必备工具列表 Visual Studio 这个似乎是不言而喻的,只是从严谨的角度,也列在这.实际上,现在也有一个 ...
- strcpy strlen memcpy等的函数实现
#include <assert.h> #include <string.h> #include <stdlib.h> #include <stdio.h&g ...
- 转:图解Git[强烈推荐]
https://my.oschina.net/xdev/blog/114383 再次感谢原著作者和中文翻译者. 此页图解git中的最常用命令.如果你稍微理解git的工作原理,这篇文章能够让你理解的更透 ...
- 转:C/C++程序员简历模板
https://github.com/geekcompany/ResumeSample/blob/master/c.md 本简历模板由国内首家互联网人才拍卖网站「 JobDeer.com 」提供. ( ...
- 2017 New Year’s Greetings from Sun Yat-sen University
As winter turns to spring, the world around us begins to take on an air of freshness. As 2017 is fa ...
- js流程控制题——如何实现一个LazyMan
先说一下想要的效果: lazyMan('zz').eat('lunch').sleep('3').eat('dinner')输出: Hi!This is zz! Eat lunch~ //有3s间隔等 ...
- (转)assert 断言式编程
编写代码时,我们总是会做出一些假设,断言就是用于在代码中捕捉这些假设,可以将断言看作是异常处理的一种高级形式.断言表示为一些布尔表达式,程序员相信在程序中的某个特定点该表达式值为真.可以在任何时候启用 ...