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先求个异或前缀和,然后就变成求k对最大异或和

因为(i^j) = (j^i)

所以我们设2k对,到答案在除以2就可以了

我们先找出每个点能得到的最大异或和,然后放堆里

每次取出堆顶,求出它能得到的第rk+1大的异或和

如果rk+1 \le n,那就放到堆里

(因为一个点最多用n次,要不就重复了)

记得long\ long

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;i++)
    #define Fdr(i,x,y) for(int i=x;i>=y;i--)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 50000011
int n,k,siz[N],c[N][2],cnt=0;
ll s[N],a[N];
void ins(ll v){
    int x=0;
    Fdr(i,31,0){
        int t=(v>>i)&1;siz[x]++;
        if(!c[x][t])c[x][t]=++cnt;
        x=c[x][t];
    }
    siz[x]++;
}
ll ask(ll v,int rk){
    int x=0;ll ans=0;
    Fdr(i,31,0){
        int t=(v>>i)&1;
        if(!c[x][!t])x=c[x][t];
        else if(rk<=siz[c[x][!t]])x=c[x][!t],ans|=1ll<<i;
        else rk-=siz[c[x][!t]],x=c[x][t];
    }
    return ans;
}
struct node{
    ll v;
    int rk,id;
    bool operator<(const node&y)const{
        return v<y.v;
    }
};
priority_queue<node>q;
int main(){
    in>>n>>k;k<<=1;
    ll ans=0,x;
    ins(s[0]);
    Fur(i,1,n)in>>x,s[i]=s[i-1]^x,ins(s[i]);
    Fur(i,0,n)q.push(node{ask(s[i],1),1,i});
    Fur(i,1,k){
        node t=q.top();q.pop();
        ans+=t.v;
        if(t.rk<n)q.push(node{ask(s[t.id],t.rk+1),t.rk+1,t.id});
    }
    out<<(ans>>1)<<ln;
}
LG 5283 [十二省联考2019]异或粽子
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