BZOJ 3223: Tyvj 1729 文艺平衡树

二次联通门 : BZOJ 3223: Tyvj 1729 文艺平衡树

 

 

 

 

/*
    BZOJ 3223: Tyvj 1729 文艺平衡树
     
    
    splay 区间翻转 
    
    每次翻转区间[l,r]都是把l旋到根节点上, r旋到根节点的右节点上
    那么要修改的区间就是根节点的右孩子的左子树
    然后打标记。。
    之后每次查到时就下放标记 
    
    原区间为1 ~ N
    增加两个哨兵节点 0 和 N + 1, 避免翻转1~N的区间产生麻烦 
     
*/ 
#include <cstdio>

#define Max 200009

inline int swap (int &a, int &b)
{
    int now = a;
    a = b;
    b = now;
}

inline void read (int &now)
{
    now = 0;
    register char word = getchar ();
    while (word < '0' || word > '9')
        word = getchar ();
    while (word >= '0' && word <= '9')
    {
        now = now * 10 + word - '0';
        word = getchar ();
    }
}

int value[Max];
int N, M;

class Splay_Tree_Type
{
    private :
        
        struct Splay_Tree_Date
        {
            int size;
            int key;
            int father;
            int child[2];
            int Flandre;
        }
        tree[Max]; 
        
        inline int Get_Son (int now)
        {
            return tree[tree[now].father].child[1] == now;
        }
        
        inline void Update (int now)
        {
            tree[now].size = 1;
            if (tree[now].child[0])
                tree[now].size += tree[tree[now].child[0]].size;
            if (tree[now].child[1])
                tree[now].size += tree[tree[now].child[1]].size;
        }
        
        int Root;
        int Answer[Max];
        int Count;
        
        inline void Down (int now)
        {
            tree[now].Flandre = 0;
            swap (tree[now].child[0], tree[now].child[1]);
            if (tree[now].child[0])
                tree[tree[now].child[0]].Flandre ^= 1;
            if (tree[now].child[1])
                tree[tree[now].child[1]].Flandre ^= 1;
        }
        
        inline void Rotate (int now)
        {
            int father = tree[now].father;
            int Grand = tree[father].father;
            int pos = Get_Son (now);
            if (tree[father].Flandre && father) // 注意这里标记下放的次序 
                Down (father);
            if (tree[now].Flandre && now)
                Down (now);
            tree[father].child[pos] = tree[now].child[pos ^ 1];
            tree[tree[father].child[pos]].father = father;
            tree[now].child[pos ^ 1] =  father;
            tree[father].father = now;
            tree[now].father = Grand;
            if (Grand)
                tree[Grand].child[tree[Grand].child[1] == father] = now;
            Update (father);
            Update (now);
        }
        
        void Get_Answer (int now) // 把整棵树进行中序遍历后的序列就是答案 
        {
            if (tree[now].Flandre)
                Down (now);
            if (tree[now].child[0])
                Get_Answer (tree[now].child[0]);
            Answer[++Count] = tree[now].key;
            if (tree[now].child[1])
                Get_Answer (tree[now].child[1]);
        }
        
        int Build (int l, int r, int father)
        {
            int now = l + r >> 1;
            tree[now].father = father;
            tree[now].key = value[now];
            if (now > l)
                tree[now].child[0] = Build (l, now - 1, now);
            if (now < r)
                tree[now].child[1] = Build (now + 1, r, now);
            Update (now);
            return now;
        }
        
    public :
        
        void Prepare ()
        {
            Root = 1;
            Root = Build(1, N + 2, 0);
        }
        
        void Splay (int now, int to)
        {
            for (int father; (father = tree[now].father) != to; Rotate (now))
                if (tree[father].father != to)
                    Rotate (Get_Son (now) == Get_Son (father) ? father : now);
            if (!to)
                Root = now;
        }
        
        int Get_Pos (int x)
        {
            int now = Root;
            for (; ; )
            {
                if (tree[now].Flandre)
                    Down (now);
                if (x <= tree[tree[now].child[0]].size)
                    now = tree[now].child[0];
                else 
                {
                    x -= tree[tree[now].child[0]].size + 1;
                    if (!x)
                        return now;
                    now = tree[now].child[1];
                }
            }
        }
        
        inline void Print ()
        {
            Get_Answer (Root);
            for (int i = 1; i <= N; i++)
                printf ("%d ", Answer[i + 1]);
        }
            
        inline void Hit_flag ()
        {
            tree[tree[tree[Root].child[1]].child[0]].Flandre ^= 1;
        }
};

Splay_Tree_Type Make;

int main (int argc, char *argv[])
{
    read (N);
    read (M); // 0号点和 N+1 号点代表是哨兵节点 
    for (int i = 1; i <= N + 2; i++) // 预先处理出值 
        value[i] = i - 1;
    int l, r;
    Make.Prepare ();  // splay建树, 要建满二叉树 
    for (; M--; )
    {
        read (l);
        read (r);
        r += 2;
        l = Make.Get_Pos (l); // 找到在树中对应的位置 
        r = Make.Get_Pos (r);   
        Make.Splay (l, 0);
        Make.Splay (r, l); 
        Make.Hit_flag ();  // 打上标记 
    }
    Make.Print (); 
    return 0;
}

 

myj 吊打我Orz,xxy 捆起来打我Orz,myl 文化课上天Orz, lrh 姿势水平敲高Orz, hkd 特别胖Orz%%%,cys 智商感人Orz,syl zz专业Orz,我没有学上, 我们未来一片光明
posted @ 2017-05-20 16:09  ZlycerQan  阅读(125)  评论(0编辑  收藏  举报