【POJ 2201 --- Cartesian Tree】

Let us consider a special type of a binary search tree, called a cartesian tree. Recall that a binary search tree is a rooted ordered binary tree, such that for its every node x the following condition is satisfied: each node in its left subtree has the key less then the key of x, and each node in its right subtree has the key greater then the key of x.
That is, if we denote left subtree of the node x by L(x), its right subtree by R(x) and its key by kx then for each node x we have
if y ∈ L(x) then ky < kx
if z ∈ R(x) then kz > kx

The binary search tree is called cartesian if its every node x in addition to the main key kx also has an auxiliary key that we will denote by ax, and for these keys the heap condition is satisfied, that is
if y is the parent of x then ay < ax

Thus a cartesian tree is a binary rooted ordered tree, such that each of its nodes has a pair of two keys (k, a) and three conditions described are satisfied.
Given a set of pairs, construct a cartesian tree out of them, or detect that it is not possible.

Input

The first line of the input file contains an integer number N – the number of pairs you should build cartesian tree out of (1 <= N <= 50 000). The following N lines contain two numbers each – given pairs (ki, ai). For each pair |ki|, |ai| <= 30 000. All main keys and all auxiliary keys are different, i.e. ki != kj and ai != aj for each i != j.

Output

On the first line of the output file print YES if it is possible to build a cartesian tree out of given pairs or NO if it is not. If the answer is positive, on the following N lines output the tree. Let nodes be numbered from 1 to N corresponding to pairs they contain as they are given in the input file. For each node output three numbers – its parent, its left child and its right child. If the node has no parent or no corresponding child, output 0 instead.
The input ensure these is only one possible tree.

Sample Input

7
5 4
2 2
3 9
0 5
1 3
6 6
4 11

Sample Output

YES
2 3 6
0 5 1
1 0 7
5 0 0
2 4 0
1 0 0
3 0 0

• 解题思路
  题目中说所有的key值都不一样，因此不会存在NO的情况。直接建树再输出即可。

1. 根据k值将节点排序。
2. 令第一个节点为根节点，从第二个节点开始遍历数组，将节点逐个插入树中：
3. 将节点插入树中的步骤为：
   从当前待插入节点的前一个节点开始，自底向上找出一个节点，使得该节点的a值比当前节点的a值小。
4. 将节点插入时有两种情况：

   1. 如果没找到：
      则当前节点为新的根节点，原来的根节点改为当前节点的左孩子。

   2. 如果找到：
      设找到的节点标号为i，则i节点原来的右孩子改为当前节点的左孩子，当前节点为节点i的右孩子。

AC代码：

#include 
#include 
#include 
using namespace std;
#define MAXN 50010
struct Node
{
    int k,a;
    int num;
}arr[MAXN];
int x[MAXN];
int father[MAXN],child_left[MAXN],child_right[MAXN];
bool compare(Node &a,Node &b)
{
    return a.k<b.k;
}

int fun(int index)
{
    int i=arr[index-1].num;
    int j;
    while(i!=0 && arr[x[i]].a>arr[index].a)
    {
        j=i;
        i=father[i];
    }

    if(i==0)
    {
        //当没有找到符合的节点时，当前节点为根节点，前一个节点为当前节点的左孩子 
        father[j]=arr[index].num;
        child_left[arr[index].num]=j;
    }
    else
    {
        //找到符合的节点时，则当前节点为i的右孩子，i节点原来的右孩子改为当前节点的左孩子 
        child_left[arr[index].num]=child_right[i];
        father[child_right[i]]=arr[index].num;
        father[arr[index].num]=i;
        child_right[i]=arr[index].num;
    }
    return 0;
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&arr[i].k,&arr[i].a);
        arr[i].num = i+1;
    }
    sort(arr,arr+n,compare);
    x[arr[0].num]=0;
    for(int i=1; i<n; i++)
    {
        x[arr[i].num]=i;
        fun(i);
    }
    printf("YES\n");
    for(int i=1;i<=n;i++)
    {
        printf("%d %d %d\n",father[i],child_left[i],child_right[i]);
    }
    return 0;
}