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 S1and 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
Greater New York 2006 一道模拟题， 解题思路：这个直接模拟（或者退化的BFS或DFS吧，就一种情况，汗。~）题目所要求的Shuffle的过程就行了，只是要注意搜索失败的条件。当S1和S2又回到初始化状态时就是搜索失败的条件，为什么会这样呢？把所有的情况放在一个集合里面，Shuffle会模拟到所有情况，直到回到初始状态，无限循环下去。
代码： #include
#include
#include
#include
#include
using namespace std; #define M 110 char mat[220],S1[M],S2[M]; char map[1000][220]; int main() { int n,m,i,j,x,k; while(scanf("%d",&m)!=EOF&&m) { for(k=1;k<=m;k++) { memset(map,'0',sizeof(map)); memset(mat,'0',sizeof(mat)); scanf("%d%s%s%s",&n,S1,S2,mat); x=1; while(1) { for(i=0;i
