UVA11212-Editing a Book(迭代加深搜索)

Problem UVA11212-Editing a Book

Accept：572  Submit：4428

Time Limit: 10000 mSec

 Problem Description

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 Ouput

Case 1: 2

Case 2: 1

题解：第一到IDA*算法的题目。迭代加深搜索，就是在普通DFS上加了个深度限制，如果目前的深度无法得到结果，就让深度限制变大，直到找到解。该算法中最重要的就是估价函数，用该函数进行剪枝操作，当前深度为d，如果最理想的情况下还要搜索h层，那么当d+h > maxd时显然就可以剪枝，其余部分和普通dfs区别不大。

续：今天对这道题又进行了一个小小的尝试，收获很大。之所以对它进行二次尝试，主要是因为第一次写时按照lrj的思路copy的，想真正自己实现一次，这次实现完全以lrj在讲解该算法时的思路为框架进行code，dfs第一步判断深度是否达到maxd，达到了就进行判断是否成立，返回信息。关键在于下一步的是先枚举还是先剪枝，如果先剪枝，两个代码的效率几乎没有差距，如果先枚举，直接TLE，这份代码用时640ms而限制是10000ms，这一个顺序使得效率差了十倍不止，仔细想了想，先剪枝与后剪枝相比，把剪枝操作提前了一层，对于每一层都有很多节点的搜索来说是很不划算的。千万注意这个顺序！！！

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>

 using namespace std;

 const int maxn = ;
 int n,order[maxn];
 int maxd;

 int cal() {
     int cnt = ;
     for (int i = ; i < n-; i++) {
         if (order[i] != order[i + ] - ) cnt++;
     }
     if (order[n - ] != n) cnt++;
     return cnt;
 }

 bool is_sorted() {
     for (int i = ; i < n - ; i++) {
         if (order[i] >= order[i + ]) return false;
     }
     return true;
 }

 bool dfs(int d) {
     if ( * d + cal() > maxd * ) return false;
     if (is_sorted()) return true;

     int old[maxn],now[maxn];
     memcpy(old, order, sizeof(order));
     for (int i = ; i < n; i++) {
         for (int j = i; j < n; j++) {
             int cnt = ;
             for (int k = ; k < n; k++) {
                 if (k < i || k > j) {
                     now[cnt++] = order[k];
                 }
             }
             for (int k = ; k <= cnt; k++) {
                 int cnt2 = ;
                 for (int p = ; p < k; p++) order[cnt2++] = now[p];
                 for (int p = i; p <= j; p++) order[cnt2++] = old[p];
                 for (int p = k; p < cnt; p++) order[cnt2++] = now[p];
                 if (dfs(d + )) return true;
                 memcpy(order, old, sizeof(order));
             }
         }
     }
     return false;
 }

 int iCase = ;

 int main()
 {
     //freopen("input.txt", "r", stdin);
     while (~scanf("%d", &n) && n) {
         for (int i = ; i < n; i++) {
             scanf("%d", &order[i]);
         }
         for (maxd = ; maxd < n; maxd++) {
             if (dfs()) break;
         }
         printf("Case %d: %d\n", iCase++, maxd);
     }
     return ;
 }