2018.08.05 bzoj3223: Tyvj 1729 文艺平衡树(非旋treap)

传送门

经典的平衡树问题,之前已经用splay写过一次了,今天我突发奇想,写了一发非旋treap的版本,发现挺好写的(虽然跑不过splay)。

代码:

#include<bits/stdc++.h>
#define N 100005
using namespace std;
typedef pair<int,int> res;
int rt=0,n,m,cnt=0,son[N][2],siz[N],val[N],rd[N],rev[N];
inline int read(){
    int ans=0;
    char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
    return ans;
}
inline int build(int v){val[++cnt]=v,rd[cnt]=rand(),siz[cnt]=1,son[cnt][0]=son[cnt][1]=0;return cnt;}
inline void pushup(int p){siz[p]=siz[son[p][0]]+siz[son[p][1]]+1;}
inline void pushdown(int p){
    if(!rev[p])return;
    swap(son[p][0],son[p][1]);
    if(son[p][0])rev[son[p][0]]^=1;
    if(son[p][1])rev[son[p][1]]^=1;
    rev[p]^=1;
}
inline int merge(int a,int b){
    if(!a||!b)return a+b;
    pushdown(a),pushdown(b);
    if(rd[a]<rd[b]){son[a][1]=merge(son[a][1],b),pushup(a);return a;}
    son[b][0]=merge(a,son[b][0]),pushup(b);return b;
}
inline res split(int p,int k){
    if(!p)return res(0,0);
    res ans,tmp;
    pushdown(p);
    if(siz[son[p][0]]>=k){
        tmp=split(son[p][0],k);
        son[p][0]=tmp.second,pushup(p);
        ans.first=tmp.first,ans.second=p;
        return ans;
    }
    tmp=split(son[p][1],k-siz[son[p][0]]-1);
    son[p][1]=tmp.first,pushup(p);
    ans.first=p,ans.second=tmp.second;
    return ans;
}
inline int rank(int p,int k){
    if(!p)return 0;
    if(val[p]>k)return rank(son[p][0],k);
    return rank(son[p][1],k)+siz[son[p][0]]+1;
}
inline int build(int l,int r){
    if(l==r){siz[l]=1,rd[l]=rand(),son[l][0]=son[l][1]=0,val[l]=l;return l;}
    int mid=l+r>>1;
    return merge(build(l,mid),build(mid+1,r));
}
inline void update(int l,int r){
    res x=split(rt,l-1),y=split(x.second,r-l+1);
    rev[y.first]^=1;
    rt=merge(x.first,merge(y.first,y.second));
}
inline void print(int p){
    pushdown(p);
    if(son[p][0])print(son[p][0]);
    cout<<p<<' ';
    if(son[p][1])print(son[p][1]);
}
int main(){
    srand(time(NULL));
    n=read(),m=read();
    rt=build(1,n);
    while(m--){
        int l=read(),r=read();
        update(l,r);
    }
    print(rt);
    return 0;
}
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