伸展树splay+uva11922

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;

struct Node
{
    Node *ch[2];
    int s;
    int flip;
    int v;
    int cmp(int k) const
    {
        int d = k - ch[0]->s;
        if(d == 1) return -1;
        return d <= 0 ? 0 : 1;
    }
    void maintain()
    {
        s = ch[0]->s + ch[1]->s + 1;
    }
    void pushdown()
    {
        if(flip)
        {
            flip = 0;
            swap(ch[0], ch[1]);
            ch[0]->flip = !ch[0]->flip;
            ch[1]->flip = !ch[1]->flip;
        }
    }
};

Node *null = new Node();

void rotate(Node* &o, int d)
{
    Node* k = o->ch[d^1];
    o->ch[d^1] = k->ch[d];
    k->ch[d] = o;
    o->maintain();
    k->maintain();
    o = k;
}

void splay(Node* &o, int k)
{
    o->pushdown();
    int d = o->cmp(k);
    if(d == 1) k -= o->ch[0]->s + 1;
    if(d != -1)
    {
        Node* p = o->ch[d];
        p->pushdown();
        int d2 = p->cmp(k);
        int k2 = (d2 == 0 ? k : k - p->ch[0]->s - 1);
        if(d2 != -1)
        {
            splay(p->ch[d2], k2);
            if(d == d2) rotate(o, d^1);
            else rotate(o->ch[d], d);
        }
        rotate(o, d^1);
    }
}

// 合并left和right。假定left的所有元素比right小。注意right可以是null，但left不可以
Node* merge(Node* left, Node* right)
{
    splay(left, left->s);
    left->ch[1] = right;
    left->maintain();
    return left;
}

// 把o的前k小结点放在left里，其他的放在right里。1<=k<=o->s。当k=o->s时，right=null
void split(Node* o, int k, Node* &left, Node* &right)
{
    splay(o, k);
    left = o;
    right = o->ch[1];
    o->ch[1] = null;
    left->maintain();
}

const int maxn = 100000 + 10;
struct SplaySequence
{
    int n;
    Node seq[maxn];
    Node *root;

    Node* build(int sz)
    {
        if(!sz) return null;
        Node* L = build(sz/2);
        Node* o = &seq[++n];
        o->v = n; // 节点编号
        o->ch[0] = L;
        o->ch[1] = build(sz - sz/2 - 1);
        o->flip = o->s = 0;
        o->maintain();
        return o;
    }

    void init(int sz)
    {
        n = 0;
        null->s = 0;
        root = build(sz);
    }
};

vector<int> ans;
void print(Node* o)
{
    if(o != null)
    {
        o->pushdown();
        print(o->ch[0]);
        ans.push_back(o->v);
        print(o->ch[1]);
    }
}


SplaySequence ss;
int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    ss.init(n+1); // 最前面有一个虚拟结点

    while (m--)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        Node *left, *mid, *right, *o;
        split(ss.root, a, left, o); // 如果没有虚拟结点，a将改成a-1，违反split的限制
        split(o, b-a+1, mid, right);
        mid->flip ^= 1;
        ss.root = merge(merge(left, right), mid);
    }

    print(ss.root);
    for(int i = 1; i < ans.size(); i++)
        printf("%d\n", ans[i]-1); // 节点编号减1才是本题的元素值

    return 0;
}