【題目描述】

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

【題目分析】
感覺這道題并不是經典的搜索，完全是一個模擬，也沒有多種情況，在網上看其他人的博客好像都是用的BFS或者是DFS，覺得不必要
關鍵還是判斷無法洗出那樣情況的判斷，發現用set很方便，將所有出現過的情況都加入到set里面，一旦重復就說明洗不出來了（這個時候還沒有出現想要洗出的牌，那么繼續洗下去只會一直重復循環）【AC代碼】

#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<queue> #include<cmath> #include<climits> #include<set>using namespace std;int n,ans; string s1,s2,s12,s;string cat(const string& s1,const string& s2) {string cnt="";for(int i=0;i<n;i++){cnt.push_back(s2[i]);cnt.push_back(s1[i]);}return cnt; }bool work() {string a,b;set<string> st;ans=1;s=cat(s1,s2);st.insert(s);while(1){//cout<<s<<" "<<s12<<endl;if(s==s12){return true;}a=s.substr(0,n);b=s.substr(n,2*n);s=cat(a,b);ans++;if(st.find(s)!=st.end())return false;st.insert(s);} }int main() {//ios::sync_with_stdio(false);//發現vjudge上用不了這個優化，不知道為什么int T;scanf("%d",&T);for(int V=1;V<=T;V++){scanf("%d",&n);cin>>s1>>s2>>s12;if(work()){printf("%d %d\n",V,ans);}else{printf("%d -1\n",V);}}return 0; }

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