LA 4794 - Sharing Chocolate dp

题意

　　有一块\(x*y\)的巧克力,问能否恰好分成n块,每块个数如下

输入格式

n

x y

a1 a2 a3 ... an

首先\(x \times y 必然要等于 \sum\limits_{i=1}^{n}a_i\)

设集合状态为S,则转移方程为

\(f(x,y,S)=(f(x,c_0,S_0)\&\& f(x,y-c_0,S_1))\|(f(c_0,y,S_0)\&\& f(x-c_0,y,S_1)) \) 分别对应横着掰和竖着掰

由于 \(x \times y = \sum\limits_{i=1}^{n}a_i\) 故可以简化为f(x,S) x为min(x,y)

 /*
 author:jxy
 lang:C/C++
 university:China,Xidian University
 **If you need to reprint,please indicate the source**
 */
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 int org[];
 int sum[<<],Max;
 int f[][<<],vis[][<<];
 int count_one(int x) //统计1的个数
 {
     x=(x&0x55555555)+(x>>&0x55555555);
     x=(x&0x33333333)+(x>>&0x33333333);
     x=(x&0x0F0F0F0F)+(x>>&0x0F0F0F0F);
     x=(x&0x00FF00FF)+(x>>&0x00FF00FF);
     x=(x&0x0000FFFF)+(x>>&0x0000FFFF);
     return x;
 }
 int dp(int x,int S)
 {
     if(vis[x][S])return f[x][S]; //记忆化搜索
     vis[x][S]=;
     int &ans=f[x][S],S1,y=sum[S]/x;
     ans=;
     if(count_one(S)==)return ans=;
     for(int S0=(S-)&S;S0;S0=(S0-)&S)
     {
         S1=S-S0;
         if(sum[S0]%x==&&dp(min(x,sum[S0]/x),S0)&&dp(min(x,sum[S1]/x),S1)) //横着掰
             return ans=;
         if(sum[S0]%y==&&dp(min(y,sum[S0]/y),S0)&&dp(min(y,sum[S1]/y),S1)) //竖着掰
             return ans=;
     }
     return ans;
 }
 int main()
 {
     int n,i;
     int x,y,C=;
     while(~scanf("%d",&n)&&n)
     {
         scanf("%d%d",&x,&y);
         for(i=;i<n;i++)
             scanf("%d",&org[i]);
         memset(vis,,sizeof(vis));
         Max=(<<n)-;
         int S;
         for(S=;S<=Max;S++)//记录每一个状态对应的巧克力块和
         {
             sum[S]=;
             for(i=;i<n;i++)
                 if(S&(<<i))sum[S]+=org[i];
         }
         int ans=;
         if(sum[Max]==x*y)ans=dp(min(x,y),Max);
         printf("Case %d: %s\n",++C,ans?"Yes":"No");
     }
 }