Uva--529（回溯，剪枝）

2014-07-19 21:06:13

题目：

  Addition Chains 

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

• a₀ = 1
• a_m = n
• a₀<a₁<a₂<...<a_m-1<a_m
• For each k ( ) there exist two (not neccessarily different) integers i and j ( ) with a_k =a_i +a_j

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 Specification 

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

Output Specification 

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

思路：首先找出链长的最小可能值：exmin = log2(n)，根据这个长度DFS，搜索过程中：如果遍历到第cur个数，那么下一个数优先从value[cur] * 2（相当于和自身加）开始。
要注意的剪枝：如果从一个数开始一直乘以2都不能达到n，那么比这个数小的数也不用考虑了！

#include<math.h>
#include<stdio.h>
int n,s[20],exmin;
bool Dfs(int cur){
    if(cur > exmin)    return false;
    if(s[cur] == n)    return true;
    else if(s[cur] > n)    return false;
    for(int i = cur; i >= 0; --i){
        s[cur + 1] = s[cur] + s[i];
        if((s[cur + 1] << (exmin - cur - 1)) < n)   break;
        if(Dfs(cur + 1))    return true;
    }
    return false;
}
int main(){
   while(scanf("%d",&n) == 1 && n){
       exmin = (int)(log(n) / log(2));
       s[1] = 1;
       while(!Dfs(1))    ++exmin;
       for(int i = 1; i <= exmin; ++i){
           if(i != 1) printf(" ");
           printf("%d",s[i]);
       }
       puts("");
   }
   return 0;
}