POJ2248 Addition Chains

Addition Chains

Description

An addition chain for n is an integer sequence with the following four properties:

• a0 = 1
• am = n
• a0 < a1 < a2 < ... < a_m-1 < am
• For each k (1<=k<=m) there exist two (not necessarily different) integers i and j (0<=i, j<=k-1) with ak=ai+aj

You are given an integer n. Your job is to construct an addition
chain for n with minimal length. If there is more than one such
sequence, any one is acceptable.

For example, <1,2,3,5> and <1,2,4,5> are both valid solutions when you are asked for an addition chain for 5.

Input

The
input will contain one or more test cases. Each test case consists of
one line containing one integer n (1<=n<=100). Input is terminated
by a value of zero (0) for n.

Output

For each test case, print one line containing the required integer sequence. Separate the numbers by one blank.

Hint: The problem is a little time-critical, so use proper break conditions where necessary to reduce the search space.

Sample Input

5
7
12
15
77
0

Sample Output

1 2 4 5
1 2 4 6 7
1 2 4 8 12
1 2 4 5 10 15
1 2 4 8 9 17 34 68 77

Source

Ulm Local 1997

【题解】

搜索层数较少，使用ID-DFS

剪枝1：ID升序枚举

剪枝2：数列降序枚举

剪枝3：保证数列递增

剪枝4：step > ID跳出

然而无论怎么剪枝，都比不上打表

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <cstring>
 5 #define min(a, b) ((a) < (b) ? (a) : (b))
 6 #define max(a, b) ((a) > (b) ? (a) : (b))
 7
 8 inline void read(int &x)
 9 {
10     x = 0;char ch = getchar(), c = ch;
11     while(ch < ‘0‘|| ch > ‘9‘)c = ch, ch = getchar();
12     while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();
13     if(c == ‘-‘)x = -x;
14 }
15
16 const int MAXN = 100 + 10;
17
18 int num[MAXN], n, b[MAXN], ID;
19
20 int dfs(int step)
21 {
22     if(num[step - 1] == n) return 1;
23     if(step > ID)return 0;
24     register int tmp;
25     for(register int i = step - 1;i >= 1;-- i)
26         for(register int j = i;j >= 1;-- j)
27         {
28             tmp = num[i] + num[j];
29             if(tmp <= num[step - 1] || b[tmp] || tmp > n)continue;
30             num[step] = tmp;
31             b[tmp] = 1;
32             if(dfs(step + 1)) return 1;
33             b[tmp] = 0;
34         }
35     return 0;
36 }
37
38 int main()
39 {
40     num[1] = 1;
41     while(scanf("%d", &n) != EOF && n)
42     {
43         for(register int i = 1;i <= n;++ i)
44         {
45             ID = i;
46             memset(b, 0, sizeof(b));
47             if(dfs(2))
48             {
49                 for(register int j = 1;j <= i;++ j)
50                     printf("%d ", num[j]);
51                 putchar(‘\n‘);
52                 break;
53             }
54         }
55     }
56     return 0;
57 } 

POJ 2248