| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7739 | Accepted: 3559 |
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
题目大意：
就是给定三个字符串，s1,s2,ss;让你判断是否能够通过 s1 和 s2 交叉来达到 ss，
ps: s1 的第一张在第一张，然后依次交叉；
解题思路：
其实我不知道为什么bin神把这道题归为搜索，我用的是集合模拟也是能做的，
重点是去重；
上代码：
/**
2015 - 09 - 16 晚上
Author: ITAK
Motto:
今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
typedef long long LL;
const int maxn = 105;
const double eps = 1e-7;
bool vis[maxn][maxn];
const int dir[4][2]= {1, 0, 0, 1, -1, 0, 0, -1};
char map[maxn][maxn];
string s1, s2, ss, str;
int main()
{
set<string> s;
int T;
scanf("%d",&T);
for(int j=1; j<=T; j++)
{
int m;
scanf("%d", &m);
cin>>s1>>s2>>ss;
int cnt = 0;
s.clear();///清屏
///很关键,没有这句的话，str[ind++] = s2[i];这样的访问str是越界的
str = ss;
while (1)
{
int ind = 0;
for (int i=0; i<m; i++)
{
str[ind++] = s2[i];
str[ind++] = s1[i];
}
cnt++;
if(str == ss)
{
printf("%d %d\n", j, cnt);
break;
}
if(s.count(str) == 1)
{
printf("%d -1\n", j);
break;
}
else
{
/**
string str1, str2 = "War and Peace";
str1.assign( str2, 4, 3 );
//str2 字符串的第4个字符位置开始赋值给str1，长度为3个字符
cout << str1 << endl; ;
**/
s.insert(str);
s1.assign(str, 0, m);//str从0位开始的m个字符
s2.assign(str, m, m);
}
}
}
return 0;
}