hihocoder 1342 Full Binary Tree Picture完全二叉树
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转自 http://www.jianshu.com/p/e37495f72cf6
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; }
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