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http://acm.hdu.edu.cn/showproblem.php?pid=3999
As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely:
1. insert a key k to a empty tree, then the tree become a tree with
only one node;
2. insert a key k to a nonempty tree, if k is less than the root ,insert
it to the left sub-tree;else insert k to the right sub-tree.
We call the order of keys we insert “the order of a tree”,your task is,given a oder of a tree, find the order of a tree with the least lexicographic order that generate the same tree.Two trees are the same if and only if they have the same shape.
There are multiple test cases in an input file. The first line of each testcase is an integer $n(n \leq 100,000)$,represent the number of nodes.The second line has n intergers,k1 to kn,represent the order of a tree.To make if more simple, k1 to kn is a sequence of 1 to n.
One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.
4
1 3 4 2
1 3 2 4
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<vector> 7 #include<map> 8 #include<set> 9 using std::cin; 10 using std::cout; 11 using std::endl; 12 using std::find; 13 using std::set; 14 using std::map; 15 using std::pair; 16 using std::vector; 17 using std::multiset; 18 using std::multimap; 19 #define sz(c) (int)(c).size() 20 #define all(c) (c).begin(), (c).end() 21 #define iter(c) decltype((c).begin()) 22 #define cls(arr,val) memset(arr,val,sizeof(arr)) 23 #define cpresent(c, e) (find(all(c), (e)) != (c).end()) 24 #define rep(i, n) for (int i = 0; i < (int)(n); i++) 25 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i) 26 #define pb(e) push_back(e) 27 #define mp(a, b) make_pair(a, b) 28 const int Max_N = 100100; 29 typedef unsigned long long ull; 30 struct Node { 31 int v; 32 Node *ch[2]; 33 inline void setc(int _v, Node *p) { 34 v = _v; 35 ch[0] = ch[1] = p; 36 } 37 }; 38 struct BST { 39 Node *root, *null, *tail; 40 Node stack[Max_N]; 41 inline void init() { 42 tail = &stack[0]; 43 null = tail++; 44 null->setc(0, NULL); 45 root = null; 46 } 47 inline Node *newNode(int v) { 48 Node *p = tail++; 49 p->setc(v, null); 50 return p; 51 } 52 inline void insert(Node *&x, int v) { 53 if (x == null) { x = newNode(v); return; } 54 insert(x->ch[v > x->v], v); 55 } 56 inline void dfs(vector<int> &res, Node *x) { 57 if (x != null) { 58 res.pb(x->v); 59 dfs(res, x->ch[0]); 60 dfs(res, x->ch[1]); 61 } 62 } 63 inline void insert(int v) { 64 insert(root, v); 65 } 66 inline void go() { 67 vector<int> res; 68 dfs(res, root); 69 int n = sz(res); 70 rep(i, n) printf("%d%c", res[i], i < n - 1 ? ‘ ‘ : ‘\n‘); 71 } 72 }bst; 73 int main() { 74 #ifdef LOCAL 75 freopen("in.txt", "r", stdin); 76 freopen("out.txt", "w+", stdout); 77 #endif 78 int n, v; 79 while (~scanf("%d", &n)) { 80 bst.init(); 81 rep(i, n) scanf("%d", &v), bst.insert(v); 82 bst.go(); 83 } 84 return 0; 85 }
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原文地址：http://www.cnblogs.com/GadyPu/p/4572796.html
