BZOJ1552: [Cerc2007]robotic sort

1552: [Cerc2007]robotic sort

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 1428  Solved: 563
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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

HINT

Source

【题解】

顺便维护一下子树最小值所在splay的节点即可，想知道节点在序列中的位置，把它splay到根看左子树即可

需要离散化让相等的数按出现顺序从小到大排

  1 /**************************************************************
  2     Problem: 1552
  3     User: 33511595
  4     Language: C++
  5     Result: Accepted
  6     Time:3360 ms
  7     Memory:21528 kb
  8 ****************************************************************/
  9  
 10 #include <iostream>
 11 #include <cstdio>
 12 #include <cstring>
 13 #include <cstdlib>
 14 #include <algorithm>
 15 #include <queue>
 16 #include <vector>
 17 #include <map>
 18 #include <string> 
 19 #include <cmath> 
 20 #define min(a, b) ((a) < (b) ? (a) : (b))
 21 #define max(a, b) ((a) > (b) ? (a) : (b))
 22 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
 23 template <class T>
 24 inline void swap(T& a, T& b)
 25 {
 26     T tmp = a;a = b;b = tmp;
 27 }
 28 inline void read(int &x)
 29 {
 30     x = 0;char ch = getchar(), c = ch;
 31     while(ch < '0' || ch > '9') c = ch, ch = getchar();
 32     while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
 33     if(c == '-') x = -x;
 34 }
 35  
 36 const int INF = 0x3f3f3f3f;
 37 const int MAXN = 500000 + 10;
 38  
 39 int ch[MAXN][2], fa[MAXN], n, root, size[MAXN], data[MAXN], mi[MAXN], tag[MAXN];
 40 void ddfs(int x)
 41 {
 42     if(!x) return;
 43     ddfs(ch[x][0]);
 44     if(x != 1 && x != n + 2) printf("%d ", data[x]);
 45     ddfs(ch[x][1]);
 46 }
 47 int son(int x){return x == ch[fa[x]][1];}
 48 void pushup(int rt)
 49 {
 50     int l = ch[rt][0], r = ch[rt][1];
 51     size[rt] = size[l] + size[r] + 1;
 52     if(mi[l] == 0) mi[rt] = mi[r];
 53     else if(mi[r] == 0) mi[rt] = mi[l];
 54     else mi[rt] = data[mi[l]] < data[mi[r]] ? mi[l] : mi[r];
 55     mi[rt] = data[mi[rt]] < data[rt] ? mi[rt] : rt;
 56 }
 57 void pushdown(int rt)
 58 {
 59     if(!tag[rt]) return;
 60     int l = ch[rt][0], r = ch[rt][1];
 61     tag[l] ^= 1, tag[r] ^= 1;
 62     swap(ch[l][0], ch[l][1]), swap(ch[r][0], ch[r][1]);
 63     tag[rt] = 0;
 64 }
 65 void rotate(int x)
 66 {
 67     int y = fa[x], z = fa[y], b = son(x), c = son(y), a = ch[x][!b];
 68     if(z) ch[z][c] = x; else root = x; fa[x] = z;
 69     if(a) fa[a] = y; ch[y][b] = a;
 70     ch[x][!b] = y, fa[y] = x;
 71     pushup(y), pushup(x);
 72 }
 73 void dfs(int x)
 74 {
 75     if(!x) return;
 76     dfs(fa[x]);
 77     pushdown(x);
 78 }
 79 void splay(int x, int i)
 80 {
 81     dfs(x);
 82     while(fa[x] != i)
 83     {
 84         int y = fa[x], z = fa[y];
 85         if(z == i) rotate(x);
 86         else
 87             if(son(x) == son(y)) rotate(y), rotate(x);
 88             else rotate(x), rotate(x); 
 89     }
 90 }
 91 int getkth(int rt, int x)
 92 {
 93     pushdown(rt);int l = ch[rt][0];
 94     if(size[l] + 1 == x) return rt;
 95     if(x < size[l] + 1) return getkth(ch[rt][0], x);
 96     return getkth(ch[rt][1], x - size[l] - 1);
 97 }
 98 void turn(int l, int r)
 99 {
100     l = getkth(root, l - 1), r = getkth(root, r + 1);
101 //  ddfs(root), putchar('\n'); 
102     splay(l, 0), splay(r, l);
103 //  ddfs(root), putchar('\n'); 
104     int now = ch[r][0];
105     tag[now] ^= 1;swap(ch[now][0], ch[now][1]);pushdown(now);
106 }
107 //查询节点k的排名 
108 int getk(int k)
109 {
110     splay(k, 0);
111     return size[ch[k][0]] + 1;
112 }
113 int ask_mi(int l, int r)
114 {
115     l = getkth(root, l - 1), r = getkth(root, r + 1);
116 //  ddfs(root), putchar('\n'); 
117     splay(l, 0), splay(r, l);
118 //  ddfs(root), putchar('\n'); 
119     return getk(mi[ch[r][0]]);
120 }
121  
122 int cnt[MAXN];
123 bool cmp(int a, int b)
124 {
125     return data[a] < data[b];
126 }
127  
128  
129 int main()
130 {
131     memset(data, 0x3f, sizeof(data));
132     read(n);
133     for(int i = 2;i <= n + 1;++ i) cnt[i] = i, read(data[i]);
134     std::stable_sort(cnt + 2, cnt + 2 + n, cmp);
135     for(int i = 2;i <= n + 1;++ i) data[cnt[i]] = i; 
136     fa[2] = 1, ch[1][1] = 2;size[1] = n + 2;mi[1] = 2;
137     for(int i = 2;i <= n + 1;++ i) 
138     { 
139         ch[i][1] = i + 1, fa[i + 1] = i;size[i] = n + 3 - i;
140         mi[i] = i;
141     } 
142     size[n + 2] = 1;mi[n + 2] = n + 2;root = 1;
143     for(int i = n + 2;i >= 1;-- i) pushup(i);
144     for(int i = 1;i < n;++ i) 
145     {
146         int tmp = ask_mi(i + 1, n + 1);
147         printf("%d ", tmp - 1);
148         turn(i + 1, tmp);
149     }
150     int tmp = ask_mi(n + 1, n + 1);
151     printf("%d\n", tmp - 1);
152     return 0;
153 }
BZOJ1552

 

posted @ 2018-01-29 08:25  嘒彼小星  阅读(217)  评论(0)  编辑  收藏  举报