LA 3621 / POJ 3134 Power Calculus (迭代加深搜索)

題目鏈接：http://poj.org/problem?id=3134

題目大意：給出x和正整數n（1<=n<=1000）,問最少需要幾次乘除法可以得到x^n。

解法：n比較小，迭代加深搜索，淺層內求出解，加上剪枝就OK了。

迭代加深搜索學習資料：

http://www.nocow.cn/index.php/Translate:USACO/More_Search_Techniques

Code：

//#pragma comment(linker, "/STACK:102400000,102400000")
//C++
//int size = 256 << 20; // 256MB
//char *p = (char*)malloc(size) + size;
//__asm__("movl %0, %%esp\n" :: "r"(p));
//G++
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<ctime>
#include<deque>
#include<cmath>
#include<vector>
#include<string>
#include<cctype>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
#define REP(i,s,t) for(int i=(s);i<=(t);i++)
#define REP2(i,s,t) for(int i=(s);i>=(t);i--)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned long ul;

int n;
const int N=1005;
int tmp[20];
int dep;
bool dfs(int cur,int x)
{
  if(tmp[cur]==n)
  {
    return 1;
  }
  if(cur>=dep)
  {
    return 0;
  }
  x=max(x,tmp[cur]);
  if(x*(1<<(dep-cur))<n)
  {
    return 0;
  }
  REP(i,0,cur)
  {
    if(x+tmp[i]<=n)
    {
      tmp[cur+1]=x+tmp[i];
      if(dfs(cur+1,x))
      {
        return 1;
      }
    }
    if(x-tmp[i]>=0)
    {
      tmp[cur+1]=x>tmp[i]?x-tmp[i]:tmp[i]-x;
      if(dfs(cur+1,x))
      {
        return 1;
      }
    }
  }
  return 0;
}
void solve(int n)
{
  if(n==1)
  {
    printf("%d\n",0);
  }
  else
  {
    tmp[0]=1;
    dep=1;
    while(1)
    {
      if(dfs(0,1))
      {
        break;
      }
      dep++;
    }
    printf("%d\n",dep);
  }
}
int main()
{
  #ifdef ONLINE_JUDGE
  #else
    freopen("test.in","r",stdin);
  #endif
  while(~scanf("%d",&n))
  {
    if(!n)break;
    solve(n);
  }
  return 0;
}