11212 Editing a Book
You have n equal-length paragraphs numbered 1 to n. Now you want to arrange them in the order of
1, 2, . . . , n. With the help of a clipboard, you can easily do this: Ctrl-X (cut) and Ctrl-V (paste) several
times. You cannot cut twice before pasting, but you can cut several contiguous paragraphs at the same
time - they’ll be pasted in order.
For example, in order to make {2, 4, 1, 5, 3, 6}, you can cut 1 and paste before 2, then cut 3 and
paste before 4. As another example, one copy and paste is enough for {3, 4, 5, 1, 2}. There are two
ways to do so: cut {3, 4, 5} and paste after {1, 2}, or cut {1, 2} and paste before {3, 4, 5}.
Input
The input consists of at most 20 test cases. Each case begins with a line containing a single integer n
(1 < n < 10), thenumber of paragraphs. The next line contains a permutation of 1, 2, 3, . . . , n. The
last case is followed by a single zero, which should not be processed.
Output
For each test case, print the case number and the minimal number of cut/paste operations.
Sample Input
6
2 4 1 5 3 6
5
3 4 5 1 2
0
Sample Output
Case 1: 2
Case 2: 1
解题思路：
1.简单分析我们可以发现，当n＝9时，最多只需要剪切八次即可完成排序。并且全排列数量9！＝362880不算很大，所以我们可以将当前排列作为状态，转化成十进制数存入set以便判重。然后逐渐增加解答树的深度（搜索最大深度）进行迭代加深搜索。
2.构造启发函数。本题可以定义一个后继错数：当前状态中，后继元素不正确的元素个数。可以证明，每一次剪切粘贴最多改变3个数的后继数，那么错数最多减少3.比如 1 2 4 3，错数是3，1 2 3 4，错数是0. 假设当前搜索到第d层，最大搜索深度为maxd，那么如果当前状态的错数 h>3*（maxd－d），则说明这个状态无解，剪枝；
3.状态转移：以长度递增的顺序，依次从每个元素开始剪切相应长度的一段，然后依次插入后继元素之后（用链表存储序列更方便剪切和插入操作）。
代码如下（关键内容有注释）：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <set> 5 #include <algorithm> 6 #include <ctime> 7 using namespace std; 8 9 #define print_time_ printf("time : %f ",double(clock())/CLOCKS_PER_SEC) 10 const int maxn=9; 11 set<int> v;//存储状态 12 int next_[maxn+2];//用链表存储当前序列 13 int n; 14 int maxd; 15 16 inline int Atoi(int *next){ //将当前序列转换成十进制数 17 int ans=0; 18 for(int i=next[0],j=0;j<n;j++,i=next[i]) 19 ans=ans*10+i; 20 return ans; 21 } 22 inline bool isvisited(int *A){//判重 23 return v.count(Atoi(A)); 24 } 25 inline void push_v(int *A){ 26 v.insert(Atoi(A)); 27 } 28 int h(int *next){//获得当前状态下的错数 29 int h=0; 30 for(int i=next[0],j=1;j<=n;j++,i=next[i]){ 31 if(j<n){ 32 if(i==n||(i!=n&&next[i]!=i+1)) 33 h++; 34 } 35 else if(i!=n) 36 h++; 37 } 38 return h; 39 } 40 int get_r(int& l,int& len){ //获得被剪切段的最右端 41 int r=l; 42 for(int i=0;i<len-1;i++) 43 r=next_[r]; 44 return r; 45 } 46 bool IDA(int d){ 47 if(d==maxd){ 48 if(h(next_)==0) 49 return true; 50 else return false; 51 } 52 int h_=h(next_); 53 if(h_>3*(maxd-d)) 54 return false; 55 for(int len=1;len<n;len++){ 56 for(int last=0,l=next_[0],j=1;j+len-1<=n;j++,last=l,l=next_[l]){ 57 58 int r=get_r(l, len); 59 60 for(int ptr=next_[r],i=j+len;i<=n;i++,ptr=next_[ptr]){ 61 62 next_[last]=next_[r]; 63 next_[r]=next_[ptr]; 64 next_[ptr]=l; 65 66 if(!isvisited(next_)){ 67 68 push_v(next_);//被访问 69 if(IDA(d+1)) 70 return true; 71 v.erase(Atoi(next_));//不要漏掉这一句！！ 72 } 73 next_[ptr]=next_[r]; 74 next_[r]=next_[last]; 75 next_[last]=l; 76 77 78 } 79 } 80 } 81 return false; 82 } 83 void init(){ 84 memset(next_, 0, sizeof next_); 85 v.clear(); 86 } 87 int main() { 88 int T=0; 89 while(scanf("%d",&n)&&n){ 90 T++; 91 init(); 92 for(int i=0,j=0;j<n;i=next_[i],j++){ 93 scanf("%d",&next_[i]); 94 } 95 96 for(maxd=0;;maxd++){ 97 v.clear(); 98 push_v(next_); 99 if(IDA(0)){ 100 printf("Case %d: %d ",T,maxd); 101 break; 102 } 103 } 104 } 105 //print_time_; 106 return 0; 107 }