UVA 11212 Editing a Book [迭代加深搜索IDA*]

　　　　　　　　　　　　　　　　　　　　　　　　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\n",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\n",T,maxd);
101                 break;
102             }
103         }
104     }
105     //print_time_;
106     return 0;
107 }