bzoj1078 "SCOI2008" oblique heap

1078: [SCOI2008] Skew heap time limit: ten Sec Memory Limit: 162 MB
Submit: 662 Solved: 380
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The inclined heap (skew heap) is a common data structure. It is also a binary tree and satisfies the same heap properties as a two-fork heap: The value of each non-root node
Is bigger than his father. Therefore, the value of the root is the smallest in the whole inclined heap. But the oblique heap does not have to be balanced, and the size of the left and right sons of each node does not have any
Regulations. In the subject, the values of each element in the oblique heap are different. The process of inserting a new element x in the oblique heap h is recursive: When H is empty or X
X becomes a new root when it is less than the root node of H, and the original root (if any) becomes the left son of X. When x is greater than the root node of H, the H-root node
Two subtrees trees are exchanged, and X (recursion) is inserted into the left subtree after the interchange. Give a slanted heap containing a node with a value of 0~n each time. Ask for a knot.
Sequence so that the oblique heap can be obtained by inserting these nodes sequentially in an empty tree. If the answer is inflexible one, the solution with the smallest dictionary order is output. The input is guaranteed to have
Solution.

Input

The first line contains an integer n. The second line contains n integers d1, d2, ..., dn, di < 100 means I is the left son of Di, di>=100 means I
It's di-100 's right son. Obviously 0 is always the root, so the input does not contain D0.

Output

Only one row, containing the n+1 integer, which is the insertion sequence with the smallest dictionary order.

Sample Input6
0 101 102 1 2Sample Output0 1 2 3 4 5 6

can and Heap , thinking good question

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#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath > #include <algorithm> #define F (I,j,n) for (int. i=j;i<=n;i++) #define D (i,j,n) for (int i=j;i>=n;i--) # Define ll long Long#define N 1005using namespace std;int n,root,cnt,fa[n],ls[n],rs[n],ans[n];inline int read () {int x=0,f= 1;char Ch=getchar (); while (ch< ' 0 ' | | Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();} while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();} return x*f;} inline void Solve () {int x=root;while (rs[x]!=-1) x=ls[x];int t=ls[x];if (t!=-1&&ls[t]==-1&&rs[t]==-1) X=t;ans[++cnt]=x;if (x==root) root=ls[x];int f=fa[x];if (f!=-1) ls[f]=ls[x],fa[ls[f]]=f;while (f!=-1) Swap (ls[f],rs[f ]), f=fa[f];} int main () {memset (fa,-1,sizeof (FA)); memset (ls,-1,sizeof (LS)); Memset (Rs,-1,sizeof (RS)); N=read (); F (i,1,n) {int x=read (); if (x<100) Ls[x]=i,fa[i]=x;else rs[x-100]=i,fa[i]=x-100;} F (I,0,n) solve ();D(i,cnt,1) printf ("%d", ans[i]); return 0;}

bzoj1078 "SCOI2008" oblique heap