poj 2248 Addition Chains(迭代加深搜索)

Addition Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5067		Accepted: 2742		Special Judge

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

解題思路：迭代加深搜索即可，注意下一個數k必定等於這個k加上之前得到的數。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int n,res[20];
int deep;

bool check(int cnt,int num)
{
	for(int i = 0; i < cnt; i++)
		for(int j = 0; j < cnt; j++)
			if(res[i] + res[j] == num)
				return true;
	return false;
}

bool dfs(int cnt,int num)
{
	if(cnt == deep)
	{
		if(num == n) return true;
		return false;
	}
	for(int i = 0; i <= cnt; i++)
	{
		num += res[i];
		if(num <= n)
		{
			res[cnt+1] = num;
			if(dfs(cnt+1,num)) return true;
		}
		num -= res[i];
	}
	return false;
}

int main()
{
	while(scanf("%d",&n),n)
	{
		if(n == 1) 
        {
            printf("1\n");continue;
        }
		res[0] = 1;
		deep = 1;
		while(true)
		{
			if(dfs(0,1)) break;
			deep++;
		}
		for(int i = 0; i <= deep; i++)
			printf("%d ",res[i]);
		printf("\n");
	}
	return 0;
}