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  • BZOJ 3689 异或之

    这个题有点像HNOI淘金的后半部分。。。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    #define maxn 100050
    using namespace std;
    int n,k,a[maxn],tree[maxn*32][2],sum[maxn*32],root,tot,s[maxn*10],top=0;
    struct status
    {
        int val,pos,k;
        status (int val,int pos,int k):val(val),pos(pos),k(k) {}
        status () {}
        friend bool operator < (const status &x,const status &y)
        {
            return x.val>y.val;
        }
    };
    priority_queue <status> q;
    void insert(int &now,int val,int bit)
    {
        if (!now) now=++tot;sum[now]++;
        if (bit==-1) return;
        int nb=(val&(1<<bit));
        insert(tree[now][(nb>0)],val,bit-1);
    }
    int ask(int now,int val,int k,int bit)
    {
        if (bit==-1) return 0;
        int nb=val&(1<<bit);
        if (!nb)
        {
            if (sum[tree[now][0]]>=k) return ask(tree[now][0],val,k,bit-1);
            else return ask(tree[now][1],val,k-sum[tree[now][0]],bit-1)+(1<<bit);
        }
        else
        {
            if (sum[tree[now][1]]>=k) return ask(tree[now][1],val,k,bit-1);
            else return ask(tree[now][0],val,k-sum[tree[now][1]],bit-1)+(1<<bit);
        }
    }
    int main()
    {
        scanf("%d%d",&n,&k);
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            insert(root,a[i],30);
        }
        for (int i=1;i<=n;i++) q.push(status(ask(root,a[i],2,30),i,2));
        for (int i=1;i<=2*k;i++)
        {
            status now=q.top();q.pop();
            s[++top]=now.val;q.push(status(ask(root,a[now.pos],now.k+1,30),now.pos,now.k+1));
        }
        for (int i=1;i<=2*k;i+=2) printf("%d ",s[i]);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ziliuziliu/p/6527548.html
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