BZOJ 3224: Tyvj 1728 普通平衡树 题解

作者: AmadeusChan | 来源:发表于2018-10-02 15:33 被阅读0次

    题目:http://www.lydsy.com/JudgeOnline/problem.php?id=3224

    思路:裸平衡树,直接模拟即可。

    代码(SBT):

    #include <cstdio>
    
    #include <algorithm>
    
    #include <cstdlib>
    
     
    
    using namespace std;
    
     
    
    struct node {
    
        node *left,*right;
    
        int key,s;
    
    };
    
     
    
    node *roof,*bank=new(node);
    
     
    
    int n;
    
     
    
    int left_ratote(node* &t){
    
        node *k=(*t).right;
    
        (*t).right=(*k).left;
    
        (*t).s=(*(*t).left).s+(*(*t).right).s+1;
    
        (*k).left=t;
    
        (*k).s=(*t).s+(*(*k).right).s+1;
    
        t=k;
    
        return 0;
    
    }
    
     
    
    int right_ratote(node* &t){
    
        node *k=(*t).left;
    
        (*t).left=(*k).right;
    
        (*t).s=(*(*t).left).s+(*(*t).right).s+1;
    
        (*k).right=t;
    
        (*k).s=(*(*k).left).s+(*t).s+1;
    
        t=k;
    
        return 0;
    
    }
    
     
    
    int maintain(node* &t){
    
        if ((*(*(*t).left).left).s>(*(*t).right).s){
    
            right_ratote(t);
    
            maintain((*t).right);
    
            maintain(t);
    
            return 0;
    
        }
    
        if ((*(*(*t).left).right).s>(*(*t).right).s){
    
            left_ratote((*t).left);
    
            right_ratote(t);
    
            maintain((*t).left);
    
            maintain((*t).right);
    
            maintain(t);
    
            return 0;
    
        }
    
        if ((*(*(*t).right).right).s>(*(*t).left).s){
    
            left_ratote(t);
    
            maintain((*t).left);
    
            maintain(t);
    
            return 0;
    
        }
    
        if ((*(*(*t).right).left).s>(*(*t).left).s){
    
            right_ratote((*t).right);
    
            left_ratote(t);
    
            maintain((*t).left);
    
            maintain((*t).right);
    
            maintain(t);
    
            return 0;
    
        }
    
        return 0;
    
    }
    
     
    
    int INSERT(int key,node* &t){
    
        if (t==bank){
    
            t=new(node);
    
            (*t).key=key;
    
            (*t).s=1;
    
            (*t).left=(*t).right=bank;
    
            return 0;
    
        }
    
        (*t).s++;
    
        if (key<=(*t).key){
    
            INSERT(key,(*t).left);
    
        } else INSERT(key,(*t).right);
    
        maintain(t);
    
        return 0;
    
    }
    
     
    
    int DELETE(int key,node* &t){
    
        if (key==(*t).key){
    
            if ((*t).left==bank){
    
                if ((*t).right==bank){
    
                    delete(t);
    
                    t=bank;
    
                    return 0;
    
                } else {
    
                    node *p=(*t).right;
    
                    delete(t);
    
                    t=p;
    
                    return 0;
    
                }
    
            } else {
    
                if ((*t).right==bank){
    
                    node *p=(*t).left;
    
                    delete(t);
    
                    t=p;
    
                    return 0;
    
                }
    
            }
    
            if (rand()%2){
    
                left_ratote(t);
    
                DELETE(key,(*t).left);
    
            } else {
    
                right_ratote(t);
    
                DELETE(key,(*t).right);
    
            }
    
        } else {
    
            if (key<(*t).key){
    
                DELETE(key,(*t).left);
    
            } else DELETE(key,(*t).right);
    
        }
    
        (*t).s=(*(*t).left).s+(*(*t).right).s+1;
    
        maintain(t);
    
        return 0;
    
    }
    
     
    
    int get_rank(int key,node *t){
    
        int rec=0;
    
        node *p=t;
    
        while (p!=bank){
    
            if ((*p).key<key){
    
                rec+=((*(*p).left).s+1);
    
                p=(*p).right;
    
            } else {
    
                p=(*p).left;
    
            }
    
        }
    
        return rec;
    
    }
    
     
    
    int get_num(int key,node *t){
    
        if ((key-1)==(*(*t).left).s){
    
            return (*t).key;
    
        }
    
        if ((key-1)<(*(*t).left).s){
    
            return get_num(key,(*t).left);
    
        } else {
    
            return get_num(key-(*(*t).left).s-1,(*t).right);
    
        }
    
    }
    
     
    
    int get_prefix(int key,node *t){
    
        if (t==bank){
    
            return -0x7fffffff;
    
        }
    
        if ((*t).key<key){
    
            return max((*t).key,get_prefix(key,(*t).right));
    
        } else {
    
            return get_prefix(key,(*t).left);
    
        }
    
    }
    
     
    
    int get_suffix(int key,node *t){
    
        if (t==bank){
    
            return 0x7fffffff;
    
        }
    
        if (key<(*t).key){
    
            return min((*t).key,get_suffix(key,(*t).left));
    
        } else {
    
            return get_suffix(key,(*t).right);
    
        }
    
    }
    
     
    
    int main(){
    
        srand(0);
    
        roof=(*bank).left=(*bank).right=bank;
    
        (*bank).s=0;
    
        scanf("%d",&n);
    
        while (n--){
    
            int x,y;
    
            scanf("%d%d",&x,&y);
    
            switch (x){
    
                case 1:
    
                    INSERT(y,roof);
    
                    break;
    
                case 2:
    
                    DELETE(y,roof);
    
                    break;
    
                case 3:
    
                    printf("%d\n",get_rank(y,roof)+1);
    
                    break;
    
                case 4:
    
                    printf("%d\n",get_num(y,roof));
    
                    break;
    
                case 5:
    
                    printf("%d\n",get_prefix(y,roof));
    
                    break;
    
                case 6:
    
                    printf("%d\n",get_suffix(y,roof));
    
                    break;
    
            }
    
        }
    
        return 0;
    
    }
    

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        本文标题:BZOJ 3224: Tyvj 1728 普通平衡树 题解

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