hihocoder 1342 Full Binary Tree Picture【完全二叉树】

转自 http://www.jianshu.com/p/e37495f72cf6

hihocoder 1342

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;
}

完整代码:

#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <string.h>
#include <ctype.h>
#include <vector>
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 N,M;
    scanf("%d%d",&N,&M);
    Node *root= new Node(0,0);
    buildTreeWithHeight(*root,N);
    while(M--)
    {
        int x1,x2,y1,y2;
        scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
        int amout = checkNodeInArea(*root,x1,y1,x2,y2);
        cout<<amout<<endl;
    }
    return 0;
}
    原文作者:demianzhang
    原文地址: https://www.cnblogs.com/demian/p/6533691.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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