UVa11212 Editing a Book (IDA*)

链接：http://acm.hust.edu.cn/vjudge/problem/23956分析：IDA*，设计一个乐观估价函数，由于每次剪切h最多减少3，因为最多有三个数字的后继数字发生了改变，DFS时用two pointers枚举要移动的数字串和移动到的位置然后递归枚举，直至不满足乐观估价函数或找到解后成功返回，从小到大枚举的深度上限maxd即为最少移动次数。

 1 #include <cstdio>
 2 #include <cstring>
 3
 4 const int maxn = 9;
 5
 6 int n, a[maxn];
 7
 8 bool is_sorted() {
 9     for (int i = 0; i < n - 1; i++)
10         if (a[i] >= a[i + 1]) return false;
11     return true;
12 }
13
14 int h() {
15     int cnt = 0;
16     for (int i = 0; i < n - 1; i++)
17         if (a[i] + 1 != a[i + 1]) cnt++;
18     if (a[n - 1] != n) cnt++;
19     return cnt;
20 }
21
22 bool dfs(int d, int maxd) {
23     if (d * 3 + h() > maxd * 3) return false;
24     if (is_sorted()) return true;
25     int b[maxn], olda[maxn];
26     memcpy(olda, a, sizeof(a));
27     for (int i = 0; i < n; i++)
28         for (int j = i; j < n; j++) {
29             int cnt = 0;
30             for (int k = 0; k < n; k++)
31                 if (k < i || k > j) b[cnt++] = olda[k];
32             for (int k = 0; k <= cnt; k++) {
33                 int cnt2 = 0;
34                 for (int p = 0; p < k; p++) a[cnt2++] = b[p];
35                 for (int p = i; p <= j; p++) a[cnt2++] = olda[p];
36                 for (int p = k; p < cnt; p++) a[cnt2++] = b[p];
37                 if (dfs(d + 1, maxd)) return true;
38             }
39         }
40     return false;
41 }
42
43 int solve() {
44     for (int maxd = 0; ; maxd++)
45         if (dfs(0, maxd)) return maxd;
46 }
47
48 int main() {
49     int kase = 0;
50     while (scanf("%d", &n) == 1 && n) {
51         for (int i = 0; i < n; i++) scanf("%d", &a[i]);
52         printf("Case %d: %d\n", ++kase, solve());
53     }
54     return 0;
55 }