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BZOJ 3678 wangxz与OJ 缩点Splay

时间:2015-03-06 16:02:32      阅读:130      评论:0      收藏:0      [点我收藏+]

标签:bzoj   splay   

题目大意

维护一个序列,支持
1. 插入一段序列,这个序列以1递增
2. 删除连续的一段序列
3. 查询位置p的数是多少。

思路

简单Splay维护就可以。但是后来好像被卡了,还有rope什么乱搞的都被卡了。于是观察这个插入的序列,他是一个很有规律的数列,但是插入之后我们却不一定查找这个序列中的数字,我们可以将这个数列当成一个节点插入Splay中去,这样每个节点可以记录lr来表示这个点所代表的序列是什么。当要使用一个节点的时候,将这个节点从一个连续的序列中分裂出来在使用。

CODE

#define _CRT_SECURE_NO_WARNINGS

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 20010
using namespace std;
#define WORKPATH (root->son[1]->son[0])

struct SplayTree{
    int l, r, size;
    SplayTree *son[2], *father;

    bool Check() {
        return father->son[1] == this;
    }
    void Combine(SplayTree *a, bool dir) {
        son[dir] = a;
        a->father = this;
    }
    void PushUp();
}none, *nil = &none, *root;
void SplayTree :: PushUp() {
    size = son[0]->size + son[1]->size + r - l + 1;
}
int cnt,asks;
int src[MAX];

void Pretreatment()
{
    nil->son[0] = nil->son[1] = nil->father = nil;
    nil->size = 0;
}

SplayTree *NewSplay(int l, int r)
{
    SplayTree *re = new SplayTree();
    re->l = l,re->r = r;
    re->size = r - l + 1;
    re->son[0] = re->son[1] = re->father = nil;
    return re;
}

SplayTree *BuildTree(int l, int r)
{
    if(l > r)   return nil;
    int mid = (l + r) >> 1; 
    SplayTree *re = NewSplay(src[mid],src[mid]);
    re->Combine(BuildTree(l, mid - 1), false);
    re->Combine(BuildTree(mid + 1, r), true);
    re->PushUp();
    return re;
}

inline void Rotate(SplayTree *a, bool dir)
{
    SplayTree *f = a->father;
    f->son[!dir] = a->son[dir];
    f->son[!dir]->father = f;
    a->son[dir] = f;
    a->father = f->father;
    f->father->son[f->Check()] = a;
    f->father = a;
    f->PushUp();
    if(root == f)   root = a;
}

inline void Splay(SplayTree *a, SplayTree *aim)
{
    while(a->father != aim) {
        if(a->father->father == aim)
            Rotate(a,!a->Check());
        else if(!a->father->Check()) {
            if(!a->Check())
                Rotate(a->father, true), Rotate(a, true);
            else    Rotate(a, false), Rotate(a, true);
        }
        else {
            if(a->Check())
                Rotate(a->father, false), Rotate(a, false);
            else    Rotate(a, true), Rotate(a, false);
        }
    }
    a->PushUp();
}

SplayTree *Find(SplayTree *a, int k)
{
    if(a->son[0]->size >= k)    return Find(a->son[0],k);
    k -= a->son[0]->size;
    if(k <= a->r - a->l + 1)    return a;
    return Find(a->son[1], k - (a->r - a->l + 1));
}

SplayTree *FindMax(SplayTree *a)
{
    if(a->son[1] == nil)    return a;
    return FindMax(a->son[1]);
}

SplayTree *FindMin(SplayTree *a)
{
    if(a->son[0] == nil)    return a;
    return FindMin(a->son[0]);
}

inline void Spilt(int k)
{
    if(!k || k == root->size)   return ;
    ++k;
    Splay(Find(root,k),nil);
    SplayTree *pred = FindMax(root->son[0]);
    SplayTree *succ = FindMin(root->son[1]);
    Splay(pred,nil);
    Splay(succ,root);
    int left = root->son[0]->size + root->r - root->l + 1;
    k -= left;
    if(k != WORKPATH->r - WORKPATH->l + 1) {
        int cnt = WORKPATH->r - WORKPATH->l + 1 - k;
        WORKPATH->Combine(NewSplay(WORKPATH->r - cnt + 1,WORKPATH->r), true);
        WORKPATH->r -= cnt;
    }
    if(k != 1) {
        WORKPATH->Combine(NewSplay(WORKPATH->l, WORKPATH->l + k - 2), false);
        WORKPATH->l += k - 1;
    }
}

inline void SplaySeg(int x,int y)
{
    Spilt(x - 1), Spilt(y + 1);
    x++, y++;
    Splay(Find(root, x - 1), nil);
    Splay(Find(root, y + 1), root);
}

int main()
{
    Pretreatment();
    cin >> cnt >> asks;
    for(int i = 1; i <= cnt; ++i)
        scanf("%d",&src[i]);
    root = BuildTree(0, cnt + 1);
    root->father = nil;
    for(int flag, x, y, z, i = 1; i <= asks; ++i) {
        scanf("%d", &flag);
        if(!flag) {
            scanf("%d%d%d", &x, &y, &z);
            Spilt(x), Spilt(x + 1);
            Splay(Find(root,x + 1), nil);
            Splay(Find(root,x + 2), root);
            root->son[1]->Combine(NewSplay(y, z), false);
            root->son[1]->PushUp();
            root->PushUp();
        }
        else if(flag == 1) {
            scanf("%d%d",&x,&y);
            SplaySeg(x,y);
            root->son[1]->son[0] = nil;
            root->son[1]->PushUp();
            root->PushUp();
        }
        else if(flag == 2) {
            scanf("%d",&x);
            Spilt(x);
            Splay(Find(root, x + 1), nil);
            printf("%d\n",root->l);
        }
    }
    return 0;
}

BZOJ 3678 wangxz与OJ 缩点Splay

标签:bzoj   splay   

原文地址:http://blog.csdn.net/jiangyuze831/article/details/44099085

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