【hdu 1890】Robotic Sort

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=1890

【题意】



给你n个数字;

i从1到n;

每次让你把第i小的数和数组的第i个元素之间这段区间内的数翻转

(第i小的数到了第i个位置);

让你输出每次操作前第i小的数的位置;

【题解】



伸展树的区间翻转以及删除操作;

维护区间的最小值;

for (int i = 1;i <＝ｎ；i++)

每次找到最小的数,然后看看它的左子树的大小->cnt;

答案就为i+cnt;

->这个数字在数组中左边的数的个数;

然后把这个数删掉;

(这样每次找最小的数,实际上就在找第i小的数字了);

伸展树里面维护这个区间内最小的数字的节点编号;

这个数的下标(因为值相同的话,下标要优先);

懒惰标记来搞区间翻转就好,只有在找最小值以及找下一个数字的时候要push_down;

splay里面不用push_down;



【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)
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0),cin.tie(0)
#define ls s[0]
#define rs s[1]

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 = 1e5+7;
const int oo = 21e8;

struct node{
    int v,idx,flag,sz;
    node *s[2],*f,*mi;
};

int n,cnt = 0;
node A[N],*root;

node *newnode(int v,int idx,node *fa){
    node *temp = A + cnt++;
    temp ->v = v,temp->idx = idx,temp->ls = temp->rs = 0,temp->mi = temp,temp->flag = 0;
    temp->f = fa;
    temp->sz = 1;
    return temp;
}

bool cmp(node *a,node *b){
    if (a->v==b->v)
        return a->idx < b->idx;
    else
        return a->v<b->v;
}

void pushup(node *o){
    o->sz = 1;
    if (o->ls) o->sz+=o->ls->sz;
    if (o->rs) o->sz+=o->rs->sz;
    o->mi = o;
    if (o->ls && cmp(o->ls->mi,o->mi)) o->mi = o->ls->mi;
    if (o->rs && cmp(o->rs->mi,o->mi)) o->mi = o->rs->mi;
}

int dir(node *o) {
    return o->f->rs == o;
}

void rotate(node *o) {
    int diro = dir(o);
    node *f = o->f;
    if (o->s[diro ^ 1]) o->s[diro ^ 1]->f = f;
    f->s[diro] = o->s[diro ^ 1];
    if (f->f) {
        if (f->f->ls == f) f->f->s[0] = o;
        else f->f->s[1] = o;
    }
    o->f = f->f;
    f->f = o;
    o->s[diro ^ 1] = f;
    pushup(f);
    pushup(o);
}

void fz(node *o){
    o->flag^=1;
    swap(o->ls,o->rs);
}

void pushdown(node *o){
    if (o->flag==0) return;
    o->flag = 0;
    if (o->ls) fz(o->ls);
    if (o->rs) fz(o->rs);
}

void splay(node *o) {
    while (o->f) {
        if (o->f->f && dir(o) == dir(o->f)) rotate(o->f);
        rotate(o);
    }
    root = o;
}

void cr(int v,int idx){
    if (!root){
        root = newnode(v,idx,NULL);
    }else{
        splay(root->rs = newnode(v,idx,root));
    }
}

void getmin(){
    node *o = root;
    while (o!=o->mi){
        pushdown(o);
        if (o->ls && o->ls->mi == o->mi)
            o = o->ls;
        else
            o = o->rs;
    }
    pushdown(o);
    splay(o);
}

void getnext(){
    node * o = root->rs;
    while (o->ls){
        pushdown(o);
        o = o->ls;
    }
    pushdown(o);
    splay(o);
}

int query(){
    getmin();
    node *o = root;
    getnext();
    if (o->ls) o->ls->f = root;
    root->ls = o->ls;
    if (o->ls) fz(o->ls);
    pushup(root);
    if (root->ls)
        return root->ls->sz;
    else
        return 0;
}

int main(){
//    Open();
    Close();
    while (cin >> n){
        if(n==0) break;
        root = 0;
        cnt = 0;
        rep1(i,1,n){
            int x;
            cin >> x;
            cr(x,i);
        }
        cr(oo,n+1);
        rep1(i,1,n){
            cout << i + query();
            if (i==n)
                cout << endl;
            else
                cout <<' ';
        }
    }
    return 0;
}