# 笛卡尔树模板

• 2018-03-03
• 0
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val 值升序的序列可以在 O(n) 的时间内借助单调栈构造出对应的笛卡尔树，具体方法如下：

1. 单调栈维护当前树的右链（下图中蓝框所示）
2. 按 val 从小到大访问每个待插入节点 u
3. 弹出栈中所有 wt 值大于 u 点 wt 值的点（以小根堆为例）
4. 将 u 点连在当前栈顶的右子树
5. 将被弹出的子树根连到 u 点的左子树

## Problem:

 Time Limit: 10000MS Memory Limit: 65536K Case Time Limit: 2000MS

Description

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```

Source

Northeastern Europe 2002, Northern Subregion

## Code: O(NlogN) [2628K, 1907MS]

```// 笛卡尔树 (2018.3.2)
// POJ 2201 Cartesian Tree
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

inline void getint(int &num){
char ch; bool neg = 0;
while(!isdigit(ch = getchar())) if(ch == '-') neg = 1;
num = ch - '0';
while(isdigit(ch = getchar())) num = num * 10 + ch - '0';
if(neg) num = -num;
}

int N, tops = 0;
int fa[50002], lc[50002], rc[50002];

struct Node{
int id, val, wt;
Node *lc, *rc, *fa;

inline bool operator < (const Node &node2) const {return val < node2.val;}
} CT[50002], *stk[50002];

inline Node* build(Node *CT, int Size){
for(register int i = 1; i <= Size; i++){
Node *o = NULL;
while(tops && CT[i].wt < stk[tops]->wt) o = stk[tops--];  // Maintain the monotonous stack
CT[i].lc = o, CT[i].fa = stk[tops];
if(o) o->fa = &CT[i];
if(stk[tops]) stk[tops]->rc = &CT[i];  // Redirect the pointers of relationship
stk[++tops] = &CT[i];
}
return stk[1];
}

int main(){
getint(N);
for(register int i = 1; i <= N; i++) getint(CT[i].val), getint(CT[i].wt), CT[i].id = i;
sort(CT + 1, CT + N + 1);  // BST indexes must be ascending
Node *root = build(CT, N);
for(register int i = 1; i <= N; i++){
fa[CT[i].id] = CT[i].fa ? CT[i].fa->id : 0;
lc[CT[i].id] = CT[i].lc ? CT[i].lc->id : 0;
rc[CT[i].id] = CT[i].rc ? CT[i].rc->id : 0;
}
puts("YES");  // Must have solution
for(register int i = 1; i <= N; i++) printf("%d %d %d\n", fa[i], lc[i], rc[i]);
return 0;
}
```

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OPEN AT 2017.12.10

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