[POJ2248] Addition Chains 迭代加深搜索

Addition Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5454		Accepted: 2923		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

Source

Ulm Local 1997

 

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提交地址 ： poj

 

搜索框架：依次搜索一位$k$， 枚举之前的$i$，$j$， 把$a[i] + a[j]$ 加到$a[k]$的位置上， 然后接着搜索；

剪枝：尽量从大到小枚举$i$，$j$让序列的数尽快逼近$n$；

为了不重复搜索，用一个$bool$数组存$a[i] + a[j]$ 是否已经被搜过；

还有一个十分厉害的剪枝，如果现在枚举到的$a[i]+a[j]$比$a[now-1]$小了，但是还没有搜到解，就直接判无解， $now$是现在搜到的位置，十分有用。

然后因为答案的深度很小， 所以一发迭代加深；

这样才能A掉...

 

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代码奉上：

//By zZhBr
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int n;
int ans;

int a[1100];

bool use[1005];
bool DFS(int stp)
{
    memset(use, 0, sizeof use);
    
    if(stp > ans)
    {
        if(a[ans] == n) return 1;
        else return 0;
    }
    
    for(register int i = stp - 1 ; i >= 1 ; i --)
    {
        for(register int j = i ; j >= 1 ; j --)
        {
            if(a[i] + a[j] > n) continue;
            if(!use[a[i] + a[j]])
            {
                if(a[i] + a[j] <= a[stp - 1]) return 0;
                use[a[i] + a[j]] = 1;
                a[stp] = a[i] + a[j];
                if(DFS(stp + 1)) return 1;
                a[stp] = 0;
                use[a[i] + a[j]] = 0;
            }
        }
    }
}

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) return 0;
        if(n == 1)
        {
            printf("1\n");
            continue;
        }
        if(n == 2)
        {
            printf("1 2\n");
            continue;
        }
        a[1] = 1;a[2] = 2;
        for(ans = 3 ; !DFS(3) ; ans ++);
        for(register int i = 1 ; i <= ans ; i ++)
        {
            printf("%d ", a[i]);
        }
        printf("\n");
        memset(a, 0, sizeof a);
    }
    return 0;
}

zZhBr