对应HDU题目：点击打开链接
Robotic SortTime Limit: 6000/2000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2854Accepted Submission(s): 1245
Problem Description
Somewhere deep in the Czech Technical University buildings, there are laboratories for examining mechanical and electrical properties of various materials. In one of yesterday’s presentations, you have seen how was one of the laboratories changed into a new multimedia lab. But there are still others, serving to their original purposes.In this task, you are to write software for a robot that handles samples in such a laboratory. Imagine there are material samples lined up on a running belt. The samples have different heights, which may cause troubles to the next processing unit. To eliminate such troubles, we need to sort the samples by their height into the ascending order.Reordering is done by a mechanical robot arm, which is able to pick up any number of consecutive samples and turn them round, such that their mutual order is reversed. In other words, one robot operation can reverse the order of samples on positions between A and B.A possible way to sort the samples is to find the position of the smallest one (P1) and reverse the order between positions 1 and P1, which causes the smallest sample to become first. Then we find the second one on position P and reverse the order between 2 and P2. Then the third sample is located etc.
The picture shows a simple example of 6 samples. The smallest one is on the 4th position, therefore, the robot arm reverses the first 4 samples. The second smallest sample is the last one, so the next robot operation will reverse the order of five samples on positions 2–6. The third step will be to reverse the samples 3–4, etc.Your task is to find the correct sequence of reversal operations that will sort the samples using the above algorithm. If there are more samples with the same height, their mutual order must be preserved: the one that was given first in the initial order must be placed before the others in the final order too.
Input
The input consists of several scenarios. Each scenario is described by two lines. The first line contains one integer number N , the number of samples, 1 ≤ N ≤ 100 000. The second line lists exactly N space-separated positive integers, they specify the heights of individual samples and their initial order.The last scenario is followed by a line containing zero.
Output
For each scenario, output one line with exactly N integers P1 , P1 , . . . PN ,separated by a space.Each Pi must be an integer (1 ≤ Pi ≤ N ) giving the position of the i-th sample just before the i-th reversal operation.Note that if a sample is already on its correct position Pi , you should output the number Pi anyway, indicating that the “interval between Pi and Pi ” (a single sample) should be reversed.
Sample Input
63 4 5 1 6 243 3 2 10
Sample Output
4 6 4 5 6 64 2 4 4
Source
2008 “Shun Yu Cup” Zhejiang Collegiate Programming Contest – Warm Up（2）
题意：
给n个数，每次将第i（1 <= i <= n）个位置到第i小的数所在位置之间的数进行翻转，输出的是第i小的数所在的位置
思路：
首先按元素升序对下标排序：
排序前
下标：1 2 3 4 5 6
元素：3 4 5 1 6 2
排序后
元素：1 2 3 4 5 6
下标：4 6 1 2 3 5
伸展数记录下标的值，初始时中序遍历是1~n（其实数组型伸展树的结点号就是下标），这样按顺序每次把排序后的下标伸展到根，然后给左子树加个翻转标志，输出sz[T] + i（sz[T]表示以T为根的子树的结点数，i从1开始），最后把根删除掉就行了（删除方法是把T（T为整棵树的根）的左子树右下角的结点翻转到左子树的根，并成为新的T，这时右子树肯定为空，接上原T的右子树）
注意数组型伸展操作splay(x, T)是可以直接定位结点位置的，所以不能直接向上伸展，因为上面可能有翻转标志没有下传，所以要先沿x走到T，并记录路径，再从T沿路径把可能有的标记下传到x后才可以做伸展。
C++可AC，G++就RE的代码：
#include <stdio.h>#include <stdlib.h>#include <string.h>#define N 100005#define nil (0x7fffffff)typedef struct{int id, val;}Node;Node a[N], t[N];int Left[N]; int Right[N]; int fa[N]; int sz[N]; //以i为根的子树的结点数bool flag[N]; int next[N]; //记录从待翻转结点到根的路径 void Push_down(int T) {if(nil == T) return;if(flag[T]){int tmp = Right[T];Right[T] = Left[T];Left[T] = tmp;flag[T] = 0;if(nil != Left[T]) flag[Left[T]] ^= 1;if(nil != Right[T]) flag[Right[T]] ^= 1;} } //初始为一条链void Init(int &T, int n) {bool k = 0;int cur, pre;for(cur = n; cur > 0; cur–){fa[cur] = Left[cur] = Right[cur] = nil;sz[cur] = 1;flag[cur] = 0;if(k){Right[cur] = pre;fa[pre] = cur;sz[cur] = sz[pre] + 1;}if(!k) k = 1;pre = cur;}T = cur + 1; } void R_rotate(const int x) {const int y = fa[x];const int z = fa[y];const int k = Right[x];int sx = sz[x], sy = sz[y], sk = 0;if(nil != k) sk = sz[k];Left[y] = k;Right[x] = y;if(nil != z){if(y == Left[z]) Left[z] = x;else Right[z] = x;}if(nil != k) fa[k] = y;fa[y] = x;fa[x] = z;sz[y] = sy – sx + sk;sz[x] = sx – sk + sz[y]; }void L_rotate(const int x) {const int y = fa[x];const int z = fa[y];const int k = Left[x];int sx = sz[x], sy = sz[y], sk = 0;if(nil != k) sk = sz[k];Right[y] = k;Left[x] = y;if(nil != z){if(y == Right[z]) Right[z] = x;else Left[z] = x;}if(nil != k) fa[k] = y;fa[y] = x;fa[x] = z;sz[y] = sy – sx + sk;sz[x] = sx – sk + sz[y]; }void Splay(int x, int &T) {if(nil == T) return;int p, end;end = fa[T];p = x;while(T != p){ //记录从x到T的路径next[fa[p]] = p;p = fa[p];}for(p = T; ; p = next[p]){ //沿路径下传翻转标记Push_down(p);if(p == x) break;}while(end != fa[x]){p = fa[x];if(end == fa[p]){ //p是根结点if(x == Left[p]) R_rotate(x);else L_rotate(x);break;}//p不是根结点if(x == Left[p]){if(p == Left[fa[p]]){R_rotate(p); //LLR_rotate(x); //LL}else{R_rotate(x); //RLL_rotate(x);}}else{if(p == Right[fa[p]]){ //RRL_rotate(p);L_rotate(x);}else{ //LRL_rotate(x);R_rotate(x);}}}T = x; } //归并排序void Merge_sort(Node *A, int l, int r, Node *T) //[l, r){if(r – l < 2) return;int m = l + (r – l) / 2;Merge_sort(A, l, m, T);Merge_sort(A, m, r, T);int p = l, i = l, q = m;while(p < m || q < r){if(q >= r || (p < m && A[p].val <= A[q].val)){T[i].val = A[p].val;T[i++].id = A[p++].id;}else{T[i].val = A[q].val;T[i++].id = A[q++].id;}}for(i = l; i < r; i++){A[i].val = T[i].val;A[i].id = T[i].id;}}void Delete(int &T){Push_down(T);int l, r, x;l = Left[T];r = Right[T];if(nil == l && nil == r){T = nil;return;}if(nil != l) fa[l] = nil;if(nil != r) fa[r] = nil;if(nil == l){ //没有左儿子T = r;return;}x = l;Push_down(x);while(nil != Right[x]){x = Right[x]; //x为T的左子树的中序最大值结点Push_down(x);}Splay(x, l); //把x伸展到左子树的根结点T = l; //把左子树的根作为整棵树的根，此时T没有右子树//接上右子树Right[T] = r;sz[T] += sz[r];//更新结点数if(nil != r) fa[r] = T;}int main(){//freopen("in.txt","r",stdin);int n, i;int T;while(scanf("%d", &n), n){T = nil;Init(T, n);for(i = 1; i <= n; i++){scanf("%d", &a[i].val);a[i].id = i;}Merge_sort(a, 1, n + 1, t);for(i = 1; i <= n; i++){Splay(a[i].id, T);if(nil != Left[T])flag[Left[T]] ^= 1; //翻转左子树printf("%d", sz[Left[T]] + i); //输出位置if(i < n) printf(" ");else printf("\n");Delete(T); //删除根结点}}return 0;}
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，就是对虚怀若谷谦虚谨慎八个字真正理解的人，
