[POJ 2248]Addition Chains

Addition Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5263		Accepted: 2829		Special Judge

Description

An addition chain for n is an integer sequence <a0, a1,a2,...,am="">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

题目大意

请构造一个尽量短的序列 A，长度为 len，一开始会给你一个正整数 n，

满足以下条件：

    A[1]=1

A[len]=n

A[i]>A[i-1]                       （len>=i>=2）

A[x]=A[i]+A[j]           （1<=i，j<x）

题解：

考虑迭代加深的ｄｆｓ
我们一开始可以算出最少需要多少个，就是答案的下界
这个怎么算呢？从１开始不断乘２，看什么时候比ｎ大，就是下界
然后将答案往上加，用ｄｆｓ判断是否可行
这样我们可以进行减枝了
如果当前的答案是ａｎｓ，当前搜索的位置是ｘ
如果 a[x]*(2^(ans-x))还比ｎ小，就可以 return 了
这样就可以搜过去了

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int n,ans[101],depth;
bool flag;
void dfs(int cur)
{
    if(flag)return;
    if(depth==cur)
    {
        if(ans[cur]==n)flag=true;
        return;
    }
    int i,j,k;
    for(i=0;i<=cur;i++)
        for(j=i;j<=cur;j++)
        if(ans[i]+ans[j]>ans[cur]&&ans[i]+ans[j]<=n)
        {
            ans[cur+1]=ans[i]+ans[j];
            int x=ans[cur+1];
            for(k=cur+1;k<=depth;k++)x*=2;
            if(x<n)continue;
            dfs(cur+1);
            if(flag)return;
        }
}
int main()
{
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        if(!n)break;
        if(n==1){printf("1\n");continue;}
        memset(ans,0,sizeof(ans));
        flag=false;
        ans[0]=1;
        depth=0;
        int x=1;
        while(1)
        {
            x*=2;
            depth++;
            if(x>=n)break;
        }
        while(1)
        {
            dfs(0);
            if(flag)break;
            depth++;
        }
        for(i=0;i<=depth;i++)
        printf("%d ",ans[i]);
        printf("\n");
    }
    return 0;
}