【hihocoder 1329】平衡树·Splay(Splay做法)

【题目链接】:http://hihocoder.com/problemset/problem/1329

【题意】

【题解】



插入操作:…,记住每次插入之后都要把它放到根节点去就好;

询问操作:对于询问x,然后找到权值为x+1的这个节点的左子树中的最大值;(如果没有这个x+1节点,则自己插入一个,之后删掉就好);

删除操作:插入两个节点l和r;然后找到小于l且最大的数字所在的节点lu,把它提到根节点所在的位置,然后再找到大于r且最小的数字所在的节点rv;把它提到根节点的右儿子所在的位置;

则根节点的右儿子的左子树就是需要删掉的->也即代表了[l..r]这个区间



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 110;

struct node
{
    int val;
    node *par,*child[2];

    node(int val): val(val),par(NULL){}
};

int n;
char s[3];
node *root;

void rotate(node* const x, int c) {
    node* const y = x->par;
    y->child[c ^ 1] = x->child[c];
    if (x->child[c] != NULL) x->child[c]->par = y;
    x->par = y->par;
    if (y->par != NULL && y->par->child[0] == y) y->par->child[0] = x;
    else if (y->par != NULL && y->par->child[1] == y) y->par->child[1] = x;
    y->par = x;
    x->child[c] = y;
}

inline bool _splay_parent(node *x, node* (&y), node* stop) {
    return (y = x->par) != stop && (x == y->child[0] || x == y->child[1]);
}

void splay(node* const x, node* const stop) {
    for (node *y, *z; _splay_parent(x, y, stop); ) {
        if (_splay_parent(y, z, stop)) {
            const int c = y == z->child[0];
            if (x == y->child[c]) rotate(x, c ^ 1), rotate(x, c);
            else rotate(y, c), rotate(x, c);
        } else {
            rotate(x, x == y->child[0]);
            break;
        }
    }
    if (stop == NULL) root = x;
}

node *cr(node *u,int val)
{
    if (u->val==val) return u;
    if (u->child[val>u->val]==NULL)
    {
        node *v = new node(val);
        v->child[0] = v->child[1] = NULL;
        v->par = u;
        u->child[val>u->val] = v;
        return v;
    }
    return cr(u->child[val>u->val],val);
}

node *cz(node *v,int val)
{
    if (v->val==val) return v;
    if (v->child[val>v->val]==NULL) return NULL;
    return cz(v->child[val>v->val],val);
}

void cr(int x)
{
    node *u = cr(root,x);
    splay(u,NULL);
}

node *get_max(node * v)
{
    if (v->child[1]==NULL)
        return v;
    return get_max(v->child[1]);
}

node *get_min(node *v)
{
    if (v->child[0]==NULL)
        return v;
    return get_min(v->child[0]);
}

void sc(int l,int r)
{

    cr(l),cr(r);

    node *u = cz(root,l);
    splay(u,NULL);

    node *lu = get_max(u->child[0]);

    node *v = cz(root,r);
    splay(v,NULL);

    node *rv = get_min(v->child[1]);

    splay(lu,NULL);
    splay(rv,lu);

    rv->child[0] = NULL;
}

int query(int x)
{
    node *v = cz(root,x+1);
    bool ju = (v==NULL);
    if (ju) v = cr(root,x+1);
    splay(v,NULL);
    int k = get_max(v->child[0])->val;
    if (ju) sc(x+1,x+1);
    return k;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use
    root = new node(-1);
    root->child[0] = root->child[1] = NULL;
    cr(21e8);
    cin >> n;
    rep1(i,1,n)
    {
        cin >> s;
        if (s[0]=='I')
        {
            int x;
            cin >> x;
            cr(x);
        }
        else
            if (s[0]=='D'){
                int l,r;
                cin >> l >> r;
                sc(l,r);
            }
            else
                if (s[0]=='Q'){
                    int x;
                    cin >> x;
                    cout << query(x) << endl;
                }
    }
    return 0;
}