BZOJ_2679_[Usaco2012 Open]Balanced Cow Subsets _meet in middle+双指针
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.
INPUT DETAILS: There are 4 cows, with milk outputs 1, 2, 3, and 4.
Sample Output
3
首先每个数的系数只可能是0,1，-1，并且1和-1都是选的状态。
用meet in middle的思想，$3^{n/2}$枚举左边和右边，把左边选或不选的状态与和挂链，右边按和排序。
枚举左边的状态，再枚举右边的和，枚举过程中左边指针单调。
然后统计答案即可。
复杂度$O(6^{n/2})$。
代码：
// luogu-judger-enable-o2
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define mr(x,y) make_pair(x,y)
#define N 100050
#define RR register
#define O2 __attribute__((optimize("-O2")))
typedef long long ll;
int n,a[25],m;
int ans;
int head[N],to[N],nxt[N],cnt,tot,t[N],vis[1<<22];
O2 struct A {
int v,S;
bool operator < (const A &x) const {
return v<x.v;
}
}b[N];
O2 inline void add(int u,int v) {
to[++cnt]=v; nxt[cnt]=head[u]; head[u]=cnt;
}
O2 void dfs(int dep,int sum,int sta) {
if(dep==m+1) {
add(sta,sum); return ;
}
dfs(dep+1,sum,sta);
dfs(dep+1,sum+a[dep],sta|(1<<(dep-1)));
dfs(dep+1,sum-a[dep],sta|(1<<(dep-1)));
}
O2 void solve(int dep,int sum,int sta) {
if(dep==n+1) {
b[++tot].v=sum; b[tot].S=sta;
return ;
}
solve(dep+1,sum,sta);
solve(dep+1,sum+a[dep],sta|(1<<(dep-1)));
solve(dep+1,sum-a[dep],sta|(1<<(dep-1)));
}
O2 int main() {
scanf("%d",&n);
m=n/2;
RR int i,j;
for(i=1;i<=n;i++) scanf("%d",&a[i]);
dfs(1,0,0);
solve(m+1,0,0);
sort(b+1,b+tot+1);
for(i=0;i<(1<<m);i++) {
t[0]=0;
for(j=head[i];j;j=nxt[j]) {
t[++t[0]]=to[j];
}
sort(t+1,t+t[0]+1);
RR int l=1,r=1;
/*for(l=1;l<=t[0];l++) {
while(r<=tot&&b[r].v<t[l]) r++;
if(r==tot+1) break;
if(b[r].v==t[l]) {
vis[i|(b[r].S)]++;
//if(vis[i|b[r].S]==1) ans++;
}
}*/
for(l=1;l<=tot;l++) {
while(r<=t[0]&&t[r]<b[l].v) r++;
if(r==t[0]+1) break;
if(t[r]==b[l].v) {
vis[i|(b[l].S)]++;
if(vis[i|(b[l].S)]==1) ans++;
}
}
}
printf("%d\n",ans-1);
}