uva 11922 Permutation Transformer 排列变换 Splay 维护序列 翻转操作

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参考:

大白书 

http://blog.csdn.net/zstu_zlj/article/details/10242605

Splay 维护的序列操作,注意lelft不能为null,所以插入一个虚拟节点

此处第k大是指序列左数第k个数

细节见注释:

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <ctime>#include <cstdlib>#include <cstring>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <vector>#include <iostream>#include <algorithm>using namespace std;#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FD(i, b, a) for(int i = (b); i >= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(a, v) memset(a, v, sizeof(a))#define PB push_back#define MP make_pairtypedef long long LL;const int INF = 0x3f3f3f3f;const int MAXN = 100010;struct Node{    Node* ch[2];    int v;    int s;    int flip;    Node(){}    Node(int v, Node* null) : v(v){        ch[0] = ch[1] = null;        s = 1; flip = 0;    }    ///比较的是位序    int cmp(int x)///    {        int ss = ch[0]->s;        if (x == ss + 1) return -1;        else return x < ss + 1 ? 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 ^= 1;            ch[1]->flip ^= 1;        }    }};///Splay维护一个序列struct Splay{    Node *root, *null;    void init()    {        null = new Node;        null->ch[0] = null->ch[1] = null;        null->v = null->s = null->flip = 0;        root = null;    }    void build(Node* &o, int n)    {        if (n >= 0)///增设虚拟节点0,split 限制 left != null !        {            o = new Node(n, null);            build(o->ch[0], n - 1);            o->maintain();//maintain        }    }    void rotate(Node* &rt, int d)    {        Node* k = rt->ch[d ^ 1]; rt->ch[d ^ 1] = k->ch[d]; k->ch[d] = rt;        rt->maintain(); k->maintain(); rt = k;//maintain    }    ///找到rt序列左数第k个元素并伸展到根    void splay(Node* &rt, int k)    {        rt->pushdown();///pushdown        int d = rt->cmp(k);        if (d != -1)        {            if (d == 1) k -= rt->ch[0]->s + 1;            Node* p = rt->ch[d];            p->pushdown();///pushdown            int d2 = p->cmp(k);            if (d2 != -1)            {                int k2 = k;                if (d2 == 1) k2 -= p->ch[0]->s + 1;                splay(p->ch[d2], k2);///!!!!!!!!!!!                if (d == d2) rotate(rt, d ^ 1);                else rotate(rt->ch[d], d);            }            rotate(rt, d ^ 1);        }    }    ///合并left和right。假设left的所有元素比right小。注意:right可以为null,left不可以    Node* merge(Node* left, Node* right)    {        splay(left, left->s);        left->ch[1] = right;        left->maintain();//maintain        return left;    }    ///把rt的前k小的节点放在left里,其它放在right里。1<= k <= rt->s.当k = o->s时,right = null    void split(Node* rt, int k, Node* &left, Node* &right)    {        splay(rt, k);        left = rt;        right = rt->ch[1];        left->ch[1] = null;///        left->maintain();//maintain    }    void solve(int a, int b)    {        Node *left, *right, *mid, *tmp_right;        split(root, a, left, tmp_right);///!!!        split(tmp_right, b - a + 1, mid, right);        mid->flip ^= 1;        root = merge(merge(left, right), mid);    }    void out(Node* &rt)    {        if (rt == null) return ;        rt->pushdown();///pushdown        out(rt->ch[0]);        if (rt->v > 0) printf("%d\n", rt->v);        out(rt->ch[1]);//        delete rt;//        rt = null;    }    void debug(Node* &o)    {        if(o == null) return ;        printf("%d(",o->v);        if(o->ch[0]!=null) printf("%d,",o->ch[0]->v);        else printf("null,");        if(o->ch[1]!=null) printf("%d",o->ch[1]->v);        else printf("null");        puts(")");        if(o->ch[0]!=null) debug(o->ch[0]);        if(o->ch[1]!=null) debug(o->ch[1]);    }}sp;int main (){    int n, m;    int x, y;    while (~scanf("%d%d", &n, &m))    {        sp.init();        sp.build(sp.root, n);        for (int i = 1; i <= m; i++)        {            scanf("%d%d", &x, &y);            sp.solve(x, y);        }        sp.out(sp.root);    }    return 0;}


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