暑假写的两个数据结构

最近翻了一下暑假的代码，发现了点有用的东西。

斜堆

      记得是个猴子打架的题目

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=100000;
int f[N+10],head[N+10],lch[N+10],rch[N+10],w[N+10];
int merge(int A,int B)
    {
        if(A==0)return B;
        if(B==0)return A;
        if(w[A]<w[B])swap(A,B);
        rch[A]=merge(rch[A],B);
        swap(rch[A],lch[A]);
        return A;
    }
int pop(int A){return merge(rch[A],lch[A]);}
int find(int x){return f[x]==x?x:find(f[x]);}
int hebing(int a,int b)
    {
        int fa=find(a),fb=find(b);f[fa]=fb;
        int pa=head[fa],pb=head[fb];
        head[fa]=pop(head[fa]);
        head[fb]=pop(head[fb]);
        w[pa]>>=1;w[pb]>>=1;
        lch[pa]=0;rch[pa]=0;
        lch[pb]=0;rch[pb]=0;
        head[fa]=merge(head[fa],pa);
        head[fb]=merge(head[fb],pb);
        head[fb]=merge(head[fa],head[fb]);
        return fb;
    }
int main()
{
    int n;scanf("%d",&n);
    memset(lch,0,sizeof(lch));
    memset(rch,0,sizeof(rch));
    for(int i=1;i<=n;i++){
        scanf("%d",&w[i]);
        f[i]=i;head[i]=i;
    }
    int m;scanf("%d",&m);
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d %d",&a,&b);
        if(find(a)!=find(b)){
            hebing(a,b);
            printf("%d
",w[head[find(a)]]);
        }
        else printf("-1
");
    }
    return 0;
}

支持删除和查询的排序二叉树树

    为了解决约瑟夫问题转化成的数据结构问题，学长教的一个很奇怪的数据结构（其实就是平衡树的一个替代品？）

#include <iostream>
#include <cstring>
#include <cstdio>
#define ll long long
using namespace std;
const int maxn=1000000;
ll lch[maxn<<2]={0},rch[maxn<<2]={0},cnt[maxn<<2]={0},key[maxn<<2]={0};
ll top=0;
bool kill[maxn<<2]={0};
int build(ll &p,ll l,ll r)
{
    if(l>r) return 0;
    p=++top;ll tmp=0;
    key[p]=(l+r)>>1;
    cnt[p]=build(lch[p],l,key[p]-1);
    tmp   =build(rch[p],key[p]+1,r);
    //cout<<p<<","<<l<<","<<r<<","<<key[p]<<","<<cnt[p]<<","<<tmp<<endl;
    return tmp+cnt[p]+1;
}
ll dfs(ll p,ll x)
{
    if(p==0)return 0;
    ll tmp;//cout<<'*'<<key[p]<<","<<x<<endl;
    if((x==cnt[p])&&(!kill[key[p]]))
      {kill[key[p]]=1;return key[p];}
    if(((x<=cnt[p])&&(!kill[key[p]])||((x<cnt[p])&&(kill[key[p]])))&&(cnt[p]>0)){
            tmp=dfs(lch[p],x);
            cnt[p]--;
    }
       else tmp=dfs(rch[p],x-cnt[p]-((!kill[key[p]])?1:0));
    /*
    tmp=(((x<=cnt[p])&&(!kill[key[p]])||((x<cnt[p])&&(kill[key[p]])))&&(cnt[p]>0))?
         (dfs(lch[p],x)):(dfs(rch[p],x-cnt[p]-((!kill[key[p]])?1:0)));
    cnt[p]-=((x<=cnt[p])?1:0);*/
    //cout<<(x<=cnt[p])<<"..."<<lch[p]<<","<<rch[p]<<","<<cnt[p]<<endl;
    return tmp;
}
long int calc(long int n,long int m,long int k)
{
    long long root;
    long int  ans=0,step=1;
    memset(kill,0,(n<<2)*sizeof(kill[0]));
    memset(rch,0,(n<<2)*sizeof(rch[0]));
    memset(lch,0,(n<<2)*sizeof(lch[0]));
    memset(cnt,0,(n<<2)*sizeof(cnt[0]));
    build(root,1,n);
    for(int i=0;i<k;i++){
        step=((step-1+m)%(n-i));
        if(step==0) step=n-i;
        ans=dfs(root,step-1);
        //cout<<(step-1+m)<<","<<step<<","<<ans<<endl;
    }
    return ans;
}
int main()
{
    long int n,m,k;
    while(cin>>n>>m>>k) cout<<calc(n,m,k)<<endl;
    return 0;
}