2016微软探星 | Full Binary Tree Pict

承接上一篇文章，题目同样来自2016微软探星夏令营在线技术笔试。这道题主要考察的是树的构建与遍历。

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题目：
时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述
Let's draw a picture of full binary tree using ASCII characters. In this picture nodes are represented by '#'. A parent node connects its left child by '/' and its right child by ''.
For the sake of aesthetic, the nodes of the same height must be painted on the same line. And the nodes must be perfectly connected by '/' and ''. Intersections or misplacements are not allowed.
For example, this is the full binary tree of height 2:

  # 
 / \
#  #

This is the full binary tree of height 3:

   #
  / \
 /   \
 #    #
/ \  / \
#  # # #

This is the full binary tree of height 4:

        # 
       / \
      /   \
     /     \
    /       \
   /         \ 
   #         # 
  / \       / \ 
 /   \     /   \ 
 #   #     #    # 
/ \ / \    / \  / \
# # # #    # #  # #

Now we build a Cartesian coordinate system for the picture. We make the root at the origin (0, 0). The positive direction of x is to the bottom and the positive direction of y is to the right.
The full binary tree of height 2 is illustrated as below.

0 
+-----+--------> y
|
0+    #(0,0)
|    / \
|   #   # 
|(2,-2) (2,2)
| 
v
x

Given the height of the tree and a rectangle area of the picture, your task is to find out the amount of nodes in such area. For example, assuming the height of tree is 2, the left-top corner of the rectangle area is at (0, 0) and the right-bottom corner is at (2, 2), then the area contains 2 nodes (the root and its right child) totally.
输入
The first line contains two integers N and M.
N is the height of the full binary tree. M is the number of queries. (1 ≤ N≤ 20, 1 ≤ M≤ 100)Each query consists of 4 integers x1, y1, x2 and y2. (x1, y1) is the left-top corner of the area and (x2,y2) is the right-bottom corner of the area.
输出
For each query output the amount of nodes in the area.

样例输入

2 3
0 0 0 0
0 0 2 2
0 -2 2 2

样例输出

1

2
3

解释：
题目描述了一种用ASCII码绘制的满二叉树，然后将树的根设置在一个特殊坐标轴的原点（0，0），坐标轴x向下为正向，y向右是正向。树的每个树枝与节点都占用1*1的大小。现在需要求在坐标轴中任意画一个矩形，里面会有多少个树的节点。例如样例输入中，对于（0，0）与（2，2）形成的矩形里面，包含有根节点和它的右叶子节点，所以输出的是2。

分析：
1、这是是一个二叉树的问题，肯定要构造树结构，为了简单，这里就声明一个Node的结构体，通过结构体指针来构建树。代码如下：

struct Node {
    Node *lchild, *rchild;
    long px, py;
    Node(long _px, long _py)
    {
        lchild = rchild = NULL;
        px = _px;
        py = _py;
    }
};

px，py是节点的坐标，lchild与rchild分别对应左右子节点。

2、接下里就是生成树，这里输入就是树的高度，我们就根据高度来生成满二叉树。生成的时候根据题目规则，我们需要注意树的树枝占位情况。通过分析我们可以得出，高度为1的节点，它一边的树枝数量是0，高度2的为1，高度3的为2，其它高度的节点树枝数量是其子节点数量的2倍加1。这样我们可以用个递归实现。代码如下：

long stickNumWithHeight(int height)
{
    if (height == 1) {
        return 0;
    }
    if (height == 2) {
        return 1;
    }
    if (height == 3) {
        return 2;
    }
    return stickNumWithHeight(height - 1) * 2 + 1;
}

void buildTreeWithHeight(Node &node, int height)
{
    if (height == 1) {
        return;
    }
    long step = stickNumWithHeight(height) + 1;
    node.lchild = new Node(node.px + step, node.py - step);
    node.rchild = new Node(node.px + step, node.py + step);
    buildTreeWithHeight(*node.lchild, height-1);
    buildTreeWithHeight(*node.rchild, height-1);
}

3、树生成过后，我们只需要对每个矩形遍历检测这棵树，就可得到在当前矩形中节点数量，代码如下：

int checkNodeInArea(Node &node, int x1, int y1, int x2, int y2)
{
    int sum = 0;
    if (node.px >= x1 && node.py >= y1 && node.px <= x2 && node.py<= y2) {
        sum += 1;
    }
    if (node.lchild != NULL) {
        sum += checkNodeInArea(*node.lchild,x1,y1,x2,y2);
    }
    if (node.rchild != NULL) {
        sum += checkNodeInArea(*node.rchild,x1,y1,x2,y2);
    }
    return sum;
}

判定结果：

完整代码：

//
//  main.cpp
//  Full Binary Tree Picture
//
//  Created by Jiao Liu on 8/5/16.
//  Copyright © 2016 ChangHong. All rights reserved.
//

#include <iostream>
#include <math.h>

using namespace std;

struct Node {
    Node *lchild, *rchild;
    long px, py;
    Node(long _px, long _py)
    {
        lchild = rchild = NULL;
        px = _px;
        py = _py;
    }
};

long stickNumWithHeight(int height)
{
    if (height == 1) {
        return 0;
    }
    if (height == 2) {
        return 1;
    }
    if (height == 3) {
        return 2;
    }
    return stickNumWithHeight(height - 1) * 2 + 1;
}

void buildTreeWithHeight(Node &node, int height)
{
    if (height == 1) {
        return;
    }
    long step = stickNumWithHeight(height) + 1;
    node.lchild = new Node(node.px + step, node.py - step);
    node.rchild = new Node(node.px + step, node.py + step);
    buildTreeWithHeight(*node.lchild, height-1);
    buildTreeWithHeight(*node.rchild, height-1);
}

int checkNodeInArea(Node &node, int x1, int y1, int x2, int y2)
{
    int sum = 0;
    if (node.px >= x1 && node.py >= y1 && node.px <= x2 && node.py<= y2) {
        sum += 1;
    }
    if (node.lchild != NULL) {
        sum += checkNodeInArea(*node.lchild,x1,y1,x2,y2);
    }
    if (node.rchild != NULL) {
        sum += checkNodeInArea(*node.rchild,x1,y1,x2,y2);
    }
    return sum;
}

int main(int argc, const char * argv[]) {
    int N,M;
    scanf("%d %d",&N,&M);
    Node *root = new Node(0,0);
    buildTreeWithHeight(*root,N);
    while (M--) {
        int x1,y1,x2,y2;
        scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
        int amout = checkNodeInArea(*root,x1,y1,x2,y2);
        cout<<amout<<endl;
    }
    return 0;
}