POJ3134--IDDFS（迭代加深dfs）

题意：http://poj.org/problem?id=3134

应该好理解
思路：

枚举层数（也就是ans）
dfs判断d到这个深度可不可以
+各种剪枝就能过
  1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
  2 #include <cstdio>//sprintf islower isupper
  3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
  4 #include <iostream>//pair
  5 #include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
  6 #include <bitset>
  7 //#include <map>
  8 //#include<unordered_map>
  9 #include <vector>
 10 #include <stack>
 11 #include <set>
 12 #include <string.h>//strstr substr strcat
 13 #include <string>
 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 15 #include <cmath>
 16 #include <deque>
 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 18 #include <vector>//emplace_back
 19 //#include <math.h>
 20 #include <cassert>
 21 #include <iomanip>
 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 25 //******************
 26 clock_t __START,__END;
 27 double __TOTALTIME;
 28 void _MS(){__START=clock();}
 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
 30 //***********************
 31 #define rint register int
 32 #define fo(a,b,c) for(rint a=b;a<=c;++a)
 33 #define fr(a,b,c) for(rint a=b;a>=c;--a)
 34 #define mem(a,b) memset(a,b,sizeof(a))
 35 #define pr printf
 36 #define sc scanf
 37 #define ls rt<<1
 38 #define rs rt<<1|1
 39 typedef pair<int,int> PII;
 40 typedef vector<int> VI;
 41 typedef unsigned long long ull;
 42 typedef long long ll;
 43 typedef double db;
 44 const db E=2.718281828;
 45 const db PI=acos(-1.0);
 46 const ll INF=(1LL<<60);
 47 const int inf=(1<<30);
 48 const db ESP=1e-9;
 49 const int mod=(int)1e9+7;
 50 const int N=(int)1e6+10;
 51 
 52 int n;
 53 bool vis[N];
 54 int a[N];
 55 bool dfs(int now,int cnt,int top)
 56 {
 57     if(cnt==top)
 58     {
 59         if(now==n)return 1;
 60         return 0;
 61     }
 62     if((now<<(top-cnt))<n)return 0;
 63     if(now==n)return 1;
 64     for(int i=1;i<=cnt;++i)
 65     {
 66         if(now-a[i]>0&&!vis[now-a[i]])
 67         {
 68             a[cnt+1]=now-a[i];
 69             vis[now-a[i]]=1;
 70             if(dfs(now-a[i],cnt+1,top))
 71             {
 72                 vis[now-a[i]]=0;
 73                 return 1;
 74             }
 75             else
 76                 vis[now-a[i]]=0;
 77         }
 78         if(!vis[now+a[i]])
 79         {
 80             a[cnt+1]=now+a[i];
 81             vis[now+a[i]]=1;
 82             if(dfs(now+a[i],cnt+1,top))
 83             {
 84                 vis[now+a[i]]=0;
 85                 return 1;
 86             }
 87             else
 88                 vis[now+a[i]]=0;
 89         }
 90     }
 91     return 0;
 92 }
 93 
 94 int main()
 95 {
 96     while(sc("%d",&n),n!=0)
 97     {
 98         int ans=1;
 99         vis[1]=1;
100         a[1]=1;
101         while(!dfs(1,1,ans))
102         {
103             ans++;
104         }
105         pr("%d
",ans-1);
106     }
107     return 0;
108 }
109 
110 /**************************************************************************************/