BZOJ2679: [Usaco2012 Open]Balanced Cow Subsets

2679: [Usaco2012 Open]Balanced Cow Subsets

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 256  Solved: 103
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Description

Farmer John's owns N cows (2 <= N <= 20), where cow i produces M(i) units of milk each day (1 <= M(i) <= 100,000,000). FJ wants to streamline the process of milking his cows every day, so he installs a brand new milking machine in his barn. Unfortunately, the machine turns out to be far too sensitive: it only works properly if the cows on the left side of the barn have the exact same total milk output as the cows on the right side of the barn! Let us call a subset of cows "balanced" if it can be partitioned into two groups having equal milk output. Since only a balanced subset of cows can make the milking machine work, FJ wonders how many subsets of his N cows are balanced. Please help him compute this quantity.

给出N（1≤N≤20）个数M(i) (1 <= M(i) <= 100,000,000)，在其中选若干个数，如果这几个数可以分成两个和相等的集合，那么方案数加1。问总方案数。

Input

 Line 1: The integer N. 

 Lines 2..1+N: Line i+1 contains M(i).

Output

* Line 1: The number of balanced subsets of cows.

Sample Input

4 1 2 3 4
INPUT DETAILS: There are 4 cows, with milk outputs 1, 2, 3, and 4.

Sample Output

3
OUTPUT DETAILS: There are three balanced subsets: the subset {1,2,3}, which can be partitioned into {1,2} and {3}, the subset {1,3,4}, which can be partitioned into {1,3} and {4}, and the subset {1,2,3,4} which can be partitioned into {1,4} and {2,3}.

HINT

 

Source

 

题解：论文(搜索问题中的meet in the middle技巧 )题

　　　此篇论文的核心思想就是：将一些可以缩小规模再合并出结果的搜索进行优化；

　　　例如：x1+x2+x3+x4+x5+x6==0（x1--6<=D) 求解的方案数

　　　　　　我们可以暴力dfs，但是时间复杂度过不了，于是我们考虑变化一下：x1+x2+x3==x4+x5+x6

　　　　　　于是我们就可以缩小规模，只需暴力出前3个，将其放入hash中，再暴力处后三个，在hash中询问即可，

　　　　　　时间复杂度由6^D变为6^(D/2)级别，获得了极大的优化！

　　　对于此题我们在dfs过程中同时记录下，每个数字的使用情况即可！

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#define N     2000005
#define mod 1000007
using namespace std;
int now[mod],Pre[N],A[N],B[N],tot;
int n,a[25],bin[25],ans;
bool vis[N];
int read()
{
    int x=0,f=1; char ch;
    while (ch=getchar(),ch<'0'||ch>'9') if (ch=='-') f=-1;
    while (x=x*10+ch-'0',ch=getchar(),ch>='0'&&ch<='9');
    return x*f;
}
void insert(int P,int PP){
    int y=abs(P)%mod;
    for (int p=now[y]; p; p=Pre[p])
    {
        if (A[p]==P && B[p]==PP) return;
    }
    ++tot; Pre[tot]=now[y]; now[y]=tot; A[tot]=P; B[tot]=PP;
}
void query(int P,int PP){
    int y=abs(P)%mod;
    for (int p=now[y]; p; p=Pre[p])
    {
        if (A[p]==P) vis[PP+B[p]]=1;
    }
}
void dfs1(int x,int P,int PP)
{
    if (x==(n/2+1)){insert(P,PP); return ;
    }
    dfs1(x+1,P+a[x-1],PP+bin[x-1]);
    dfs1(x+1,P-a[x-1],PP+bin[x-1]);
    dfs1(x+1,P,PP);
}
void dfs2(int x,int P,int PP)
{
    if (x==(n+1)){query(P,PP); return ;
    }
    dfs2(x+1,P+a[x-1],PP+bin[x-1]);
    dfs2(x+1,P-a[x-1],PP+bin[x-1]);
    dfs2(x+1,P,PP);
}
int main()
{
    bin[0]=1; for (int i=1;i<21;i++) bin[i]=bin[i-1]<<1;
    n=read(); for (int i=0; i<n; i++) a[i]=read();
    dfs1(1,0,0); dfs2(n/2+1,0,0); 
    for (int i=1; i<(1<<n); i++) if (vis[i]) ans++;
    printf("%d\n",ans);
    return 0;
}