LA 4794(Sharing Chocolate-记忆化搜索)

一个R*C的矩形,每次操作能通过一条线切割成两个长宽均为整数的矩形
是否能通过切割,得到面积恰为a1,a2,...,an

记忆化搜索1..n的子集
dpr,S 表示当前矩形面积为S集合的总和,矩形最小边为r
注意面积和恰为R*C的条件

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
#define
int r,c,a[MAXN];
int sum[1<<MAXN], f[MAXW][1<<MAXN];
bool vis[MAXW][1<<MAXN];
int bitcount(int x) { return x==0 ? 0 : bitcount(x/2) + (x&1) ;}
int dp(int r, int S) {
    if (vis[r][S]) return f[r][S];
    vis[r][S]=1;
    f[r][S]=0;
    if (bitcount(S) == 1) return f[r][S] = 1; 
    int c = sum[S] / r;
    for(int S0 = S; S0; S0=(S0-1)&S){
        int S1 = S - S0;
        if (sum[S0]%r == 0) {
            if (dp(min(r,sum[S0]/r),S0) && dp(min(r,sum[S1]/r),S1) ) return f[r][S] = 1; 
        } 
        if (sum[S0]%c == 0) {
            if (dp(min(c,sum[S0]/c),S0) && dp(min(c,sum[S1]/c),S1) ) return f[r][S] = 1; 
        } 

    }
    return f[r][S] = 0;
}
int main()
{
//  freopen("la4794.in","r",stdin);
//  freopen(".out","w",stdout);

    int n,kcase=1;
    while(n=read()) {
        MEM(vis)
        r=read(),c=read();
        int all = (1<<n) - 1;

        Rep(i,n) {
            a[i]=read();
        }
        Rep(S,all+1) {
            sum[S]=0;
            Rep(i,n) if (S & (1<<i) ) sum[S] += a[i];
        }

        if (sum[all] != r*c || sum[all] % r != 0  ) printf("Case %d: No\n",kcase++);
        else if (dp(min(r,c),all)) printf("Case %d: Yes\n",kcase++);
        else printf("Case %d: No\n",kcase++);



    }   
    return 0;
}