【hihocoder 1333】平衡树·Splay2

【题目链接】:http://hihocoder.com/problemset/problem/1333

【题意】

【题解】



伸展树;

要求提供操作:

1.插入一个元素，两个权值,id作为查找的比较权值,val作为储存信息;

2.将id在[a..b]范围内的点的val值改变d;d能为负值

3.将id在[a..b]范围内的点全部删掉;

4.查询id在[a..b]范围内的点的val值的和;

区间的更改要用到懒惰标记;

权值和,则需要在node域里面加一个变量,存当前子树的val和;

在旋转,插入,删除的时候,要更新权值和,同时下传懒惰标记.



【Number Of WA】



14(懒惰标记往下传的时候,乘错了,应该是乘子树的大小的)



【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 id;
    LL val,sumval,lazy_tag,siz;
    node *par,*child[2];
    node(int id,LL val): id(id),val(val),sumval(val),lazy_tag(0),siz(1),par(NULL){}
};

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

void up_data(node *v)//更新这个节点,更新它的大小和权值和
{
    v->sumval = v->val,v->siz = 1;
    rep1(i,0,1)
        if (v->child[i]!=NULL)
        {
            node *y = v->child[i];
            v->sumval+=y->sumval;
            v->siz+=y->siz;
        }
}

void add_tag(node *x)//懒惰标记下传
{
    if (x->lazy_tag==0) return;
    rep1(i,0,1)
        if (x->child[i]!=NULL)
        {
            node *v = x->child[i];
            v->val+=x->lazy_tag;
            v->sumval+=1LL*x->lazy_tag*v->siz;//注意乘的是儿子节点的子树大小
            v->lazy_tag+=x->lazy_tag;
        }
    x->lazy_tag = 0;
}

void rotate(node* const x, int c) {//注意懒惰标记的下传位置
    node* const y = x->par;
    add_tag(y),add_tag(x);
    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;
    up_data(y),up_data(x);
}

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 id,int val)//返回插入的节点所在的位置,如果已经存在则返回那个节点
{
    add_tag(u);
    if (u->id==id) return u;
    if (u->child[id>u->id]==NULL)
    {
        node *v = new node(id,val);
        v->child[0] = v->child[1] = NULL;
        v->par = u;
        u->child[id>u->id] = v;
        return v;
    }
    node *temp = cr(u->child[id>u->id],id,val);
    up_data(u);
    return temp;
}

void cr(int id,int val)//插入某个节点
{
    node *u = cr(root,id,val);
    splay(u,NULL);
}

node * cz(node *u,int id)//查找某个权值,不存在就返回空
{
    if (u->id == id) return u;
    if (u->child[id>u->id]==NULL) return NULL;
    return cz(u->child[id>u->id],id);
}

node *get_max(node * v)//获取子树最大的权值所在节点
{
    if (v->child[1]==NULL)
        return v;
    return get_max(v->child[1]);
}

node *get_min(node *v)//获取子树最小的权值所在节点
{
    if (v->child[0]==NULL)
        return v;
    return get_min(v->child[0]);
}

void sc(int l,int r)//删除[l..r]这个区间
{
    cr(l,0),cr(r,0);//删除操作获取l..r的时候,是先找到l的前驱和r的后继,然后一起删掉

    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;
    up_data(rv),up_data(lu);
}

void gb(int a,int b,int c)//把区间[l..r]的值都改变c
{
    node *u = cz(root,a-1),*v = cz(root,b+1);//同理插入一个a-1和一个b+1
    bool ju1 = (u==NULL),ju2 = (v==NULL);
    if (ju1) u = cr(root,a-1,0);
    if (ju2) v = cr(root,b+1,0);
    splay(u,NULL);
    splay(v,u);
    if (v->child[0]!=NULL)
    {
        node *y = v->child[0];
        y->val+=c;
        y->sumval+=1LL*y->siz*c;
        y->lazy_tag+=c;
        up_data(v),up_data(u);
    }
    if (ju1) sc(a-1,a-1);
    if (ju2) sc(b+1,b+1);
}

LL query(int a,int b)//询问区间[a..b]的权值和
{
    node *u = cz(root,a-1),*v = cz(root,b+1);
    bool ju1 = (u==NULL),ju2 = (v==NULL);//看看有没有a-1和b+1,如果有的话,就能方便获取[a..b]了
    if (ju1) u = cr(root,a-1,0);//没有的话就自己插入一个
    if (ju2) v = cr(root,b+1,0);
    splay(u,NULL);
    splay(v,u);
    LL temp = v->child[0]->sumval;
    if (ju1) sc(a-1,a-1);
    if (ju2) sc(b+1,b+1);
    return temp;
}

int main()
{
    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use
    root = new node(-1,0);
    root->child[0] = root->child[1] = NULL;
    cr(21e8,0);
    cin >> n;
    rep1(i,1,n)
    {
        cin >> s;
        if (s[0]=='I')
        {
            int id,val;
            cin >> id >> val;
            cr(id,val);
        }
        if (s[0]=='D')
        {
            int a,b;
            cin >> a >> b;
            sc(a,b);
        }
        if (s[0]=='M')
        {
            int a,b,c;
            cin >> a >> b >> c;
            gb(a,b,c);
        }
        if (s[0]=='Q')
        {
            int a,b;
            cin >> a >> b;
            cout << query(a,b) << endl;
        }

    }
    return 0;
}