[容斥] Topcoder SRM div1-3 12004. SetAndSet

把所有数取反，转换成分成两个集合，集合或值相同。

容斥一下

每次枚举哪些位不满足条件，用并查集维护一下要在同一个集合的联通块就好了

#include  
#include 
#include 
#include 
#include 

using namespace std;


class SetAndSet{
    public:
    typedef long long ll;
    vector<int> b[25];
    int n,f[55],a[25],vis[55];
    ll ans;

    int Gfat(int x){
        return f[x]==x?x:f[x]=Gfat(f[x]);
    }

    void dfs(int x,int y){
        if(x>=20){
            int cnt=0;
            for(int i=1;i<=n;i++) vis[i]=0;
            for(int i=1;i<=n;i++){
                if(!vis[Gfat(i)]) cnt++,vis[Gfat(i)]=1;
            }
            if(y&1) ans-=(1LL<2; else ans+=(1LL<2;
            return ;
        }
        dfs(x+1,y); 
        if(b[x].size()<1) return ;
        int curf[55];
        memcpy(curf,f,sizeof(curf));
//      for(int i=1;i<=n;i++) curf[i]=f[i];
        int nf=Gfat(b[x][0]);
        for(int j=1;j1,y+1);
        memcpy(f,curf,sizeof(f));
//      for(int i=1;i<=n;i++) f[i]=curf[i];
    }

    ll countandset(vector<int> A){
        n=A.size(); ans=0;
        for(int i=1;i<=n;i++) f[i]=i;
        for(int i=0;ifor(int j=0;j<20;j++)
                if(~A[i]>>j&1) b[j].push_back(i+1);
        dfs(0,0);
        for(int i=0;i<20;i++) b[i].clear();
        return ans;
    }
};

inline void run_(int test_){
    vector<int> a; SetAndSet A;
    if(test_==0){ a.push_back(1); a.push_back(2); cout<if(test_==1){ a.push_back(1); a.push_back(2); a.push_back(3); a.push_back(4); cout<if(test_==2){ for(int i=1;i<=5;i++) a.push_back(i); cout<if(test_==3){ for(int i=1;i<=3;i++) a.push_back(6); cout<if(test_==4){ a.push_back(13); a.push_back(10); a.push_back(4); a.push_back(15); a.push_back(4); a.push_back(8); a.push_back(4); a.push_back(2); a.push_back(4); a.push_back(14); a.push_back(0); cout<if(test_==5){
        freopen("1.in","r",stdin);
        freopen("1.out","w",stdout);
        int n;scanf("%d",&n);
        for(int i=1,x;i<=n;i++)
            scanf("%d",&x),a.push_back(x);
        cout<