1552: [Cerc2007]robotic sort

Posted 2020-10-02 HWIM

1552: [Cerc2007]robotic sort

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 1205  Solved: 459

Description

Input

输入共两行，第一行为一个整数N，N表示物品的个数，1<=N<=100000。

第二行为N个用空格隔开的正整数，表示N个物品最初排列的编号。

Output

输出共一行，N个用空格隔开的正整数P1,P2,P3…Pn，Pi表示第i次操作前第i小的物品所在的位置。 

注意：如果第i次操作前，第i小的物品己经在正确的位置Pi上，我们将区间[Pi,Pi]反转(单个物品)。

Sample Input

6
3 4 5 1 6 2

Sample Output

4 6 4 5 6 6

code

 注意输出格式。

  1 #include<cstdio>
  2 #include<algorithm>
  3 
  4 using namespace std;
  5 const int MAXN = 100100;
  6 const int INF = 0x7fffffff;
  7 struct DATA{
  8     int x,p;
  9     bool operator < (const DATA &a) const 
 10     {
 11         if (x==a.x) return p < a.p;
 12         return x < a.x; 
 13     }
 14 }d[MAXN];
 15 int mn[MAXN],mnpos[MAXN],data[MAXN],siz[MAXN],fa[MAXN],ch[MAXN][2],tag[MAXN];
 16 int n,root;
 17 
 18 int read()
 19 {
 20     int x = 0,f = 1;char ch = getchar();
 21     while (ch<‘0‘||ch>‘9‘) {if(ch==‘-‘)f=-1; ch=getchar(); }
 22     while (ch>=‘0‘&&ch<=‘9‘) {x=x*10+ch-‘0‘; ch=getchar();}
 23     return x;
 24 }
 25 void pushup(int x)
 26 {
 27     int l = ch[x][0],r = ch[x][1];
 28     mn[x] = min(data[x],min(mn[l],mn[r]));
 29     if (mn[x]==data[x]) mnpos[x] = x;
 30     else if (mn[x]==mn[l]) mnpos[x] = mnpos[l];
 31     else mnpos[x] = mnpos[r];
 32     siz[x] = siz[l]+siz[r]+1;
 33 }
 34 void pushdown(int x)
 35 {
 36     if (tag[x])
 37     {
 38         tag[x] = 0;
 39         int l = ch[x][0],r = ch[x][1];
 40         tag[l] ^= 1;
 41         swap(ch[l][0],ch[l][1]);
 42         tag[r] ^= 1;
 43         swap(ch[r][0],ch[r][1]);
 44     }
 45 }
 46 int son(int x)
 47 {
 48     return ch[fa[x]][1]==x;
 49 }
 50 void rotate(int x)
 51 {
 52     int y = fa[x],z = fa[y],b = son(x),c = son(y),a = ch[x][!b];
 53     if (z) ch[z][c] = x;else root = x;fa[x] = z;
 54     if (a) fa[a] = y;ch[y][b] = a;
 55     ch[x][!b] = y;fa[y] = x;
 56     pushup(y);pushup(x);
 57 }
 58 void splay(int &x,int rt)
 59 {
 60     while (fa[x]!=rt)
 61     {
 62         int y = fa[x],z = fa[y];
 63         if (z==rt) rotate(x);
 64         else
 65         {
 66             pushdown(z);
 67             pushdown(y);
 68             pushdown(x);
 69             if (son(x)==son(y)) rotate(y),rotate(x);
 70             else rotate(x), rotate(x);
 71         }
 72     }
 73 }
 74 int getkth(int rt,int k)
 75 {
 76     pushdown(rt);
 77     int l = ch[rt][0],r = ch[rt][1];
 78     if (k==siz[l]+1) return rt;
 79     else if (k<siz[l]+1) return getkth(l,k);
 80     else return getkth(r,k-siz[l]-1);
 81 }
 82 int getmnpos(int l,int r)
 83 {
 84     int tl = getkth(root,l-1);
 85     int tr = getkth(root,r+1);
 86     splay(tl,0);
 87     splay(tr,tl);
 88     return mnpos[ch[tr][0]];
 89 }
 90 void reverse(int l,int r)
 91 {
 92     int tl = getkth(root,l-1);//根节点是root 
 93     int tr = getkth(root,r+1);
 94     splay(tl,0);
 95     splay(tr,tl);
 96     int p = ch[tr][0];
 97     tag[p] ^= 1;
 98     swap(ch[p][0],ch[p][1]);
 99 }
100 int main()
101 {
102     n = read();
103     for (int i=2; i<=n+1; ++i)
104     {
105         data[i] = read();
106         d[i].x = data[i];
107         d[i].p = i;
108     }
109     sort(d+2,d+n+2);
110     for (int i=2; i<=n+1; ++i)
111         data[d[i].p] = i;
112     for (int i=0; i<=n+2; ++i)
113         mn[i] = INF;
114     data[0] = data[1] = data[n+2] = INF;
115     siz[0] = 0;root = 1;
116     for (int i=1; i<=n+2; ++i)
117     {
118         fa[i] = i-1;
119         ch[i][1] = i+1;
120     }
121     ch[n+2][1] = 0;
122     for (int i=1; i<=n+2; ++i)
123         pushup(i);
124     for (int i=1; i<=n; ++i)
125     {
126         int p = getmnpos(i+1,n+1);
127         splay(p,0);
128         printf("%d",siz[ch[p][0]]);
129         reverse(i+1,siz[ch[p][0]]+1);
130         if (i!=n) printf(" ");
131     }
132     return 0;
133 }