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 = 9;
int n,order[maxn];
int maxd;

int cal() {
    int cnt = 0;
    for (int i = 0; i < n-1; i++) {
        if (order[i] != order[i + 1] - 1) cnt++;
    }
    if (order[n - 1] != n) cnt++;
    return cnt;
}

bool is_sorted() {
    for (int i = 0; i < n - 1; i++) {
        if (order[i] >= order[i + 1]) return false;
    }
    return true;
}

bool dfs(int d) {
    if (3 * d + cal() > maxd * 3) return false;
    if (is_sorted()) return true;

    int old[maxn],now[maxn];
    memcpy(old, order, sizeof(order));
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int cnt = 0;
            for (int k = 0; k < n; k++) {
                if (k < i || k > j) {
                    now[cnt++] = order[k];
                }
            }
            for (int k = 0; k <= cnt; k++) {
                int cnt2 = 0;
                for (int p = 0; 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 + 1)) return true;
                memcpy(order, old, sizeof(order));
            }
        }
    }
    return false;
}

int iCase = 1;

int main()
{
    //freopen("input.txt", "r", stdin);
    while (~scanf("%d", &n) && n) {
        for (int i = 0; i < n; i++) {
            scanf("%d", &order[i]);
        }
        for (maxd = 0; maxd < n; maxd++) {
            if (dfs(0)) break;
        }
        printf("Case %d: %d
", iCase++, maxd);
    }
    return 0;
}