POJ 3437 Tree Grafting(二叉树高度)



Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 1677 Accepted: 718

Description

Trees have many applications in computer science. Perhaps the most commonly used trees are rooted binary trees, but there are other types of rooted trees that may be useful as well. One example is ordered trees, in which the subtrees for any given node are ordered. The number of children of each node is variable, and there is no limit on the number. Formally, an ordered tree consists of a finite set of nodes T such that

• there is one node designated as the root, denoted root(T);
  
• the remaining nodes are partitioned into subsets T1, T2, ..., Tm, each of which is also a tree (subtrees).

Also, define root(T1), ..., root(Tm) to be the children of root(T), with root(Ti) being the i-th child. The nodes root(T1), ..., root(Tm) are siblings.

It is often more convenient to represent an ordered tree as a rooted binary tree, so that each node can be stored in the same amount of memory. The conversion is performed by the following steps:

1. remove all edges from each node to its children;
2. for each node, add an edge to its first child in T (if any) as the left child;
3. for each node, add an edge to its next sibling in T (if any) as the right child.

This is illustrated by the following:

         0                             0       / | \                          /      1  2  3       ===>             1        / \                           \       4   5                           2                                      / \                                     4   3                                      \                                       5

In most cases, the height of the tree (the number of edges in the longest root-to-leaf path) increases after the conversion. This is undesirable because the complexity of many algorithms on trees depends on its height.

You are asked to write a program that computes the height of the tree before and after the conversion.

Input

The input is given by a number of lines giving the directions taken in a depth-first traversal of the trees. There is one line for each tree. For example, the tree above would give dudduduudu, meaning 0 down to 1, 1 up to 0, 0 down to 2, etc. The input is terminated by a line whose first character is #. You may assume that each tree has at least 2 and no more than 10000 nodes.

Output

For each tree, print the heights of the tree before and after the conversion specified above. Use the format:

 Tree t: h1 => h2 

where t is the case number (starting from 1), h1 is the height of the tree before the conversion, and h2 is the height of the tree after the conversion.

Sample Input

dudduduududdddduuuuudddduduuuudddduuduuu#

Sample Output

Tree 1: 2 => 4Tree 2: 5 => 5Tree 3: 4 => 5Tree 4: 4 => 4

Source

Rocky Mountain 2007

首先,贴上多叉树转二叉树的方法:

方法非常简单,"左儿子,右兄弟"
就是将一个节点的第一个儿子放在左儿子的位置,下一个的儿子,即左儿子的第一个兄弟,
放在左儿子的右儿子位置上,再下一个兄弟接着放在右儿子的右儿子.

由题目给出的字符串,可以轻松求出多叉树.但是题目问的是转为二叉树的高度......

由于多叉树转二叉树的方法是"左儿子右兄弟",因此 二叉树中X节点的高=二叉树中X的父节点的高+X是第几个儿子-1

形成了一个递归定义. 可以用递归算法解决.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;char str[1000000];      //可以用stringint height1,height2;int len,i;void caculate(int level1,int level2){    if(i == len) return ;    int tempson = 0;    while(str[i] == 'd')    {        i++;        tempson++;        caculate(level1+1,level2+tempson);    }    if(height1 < level1)        height1 = level1;    if(height2 < level2)        height2 = level2;    i++;}int main(){    int t = 1;    while(cin>>str)    {        if(str[0] == '#')            break;        height1 = height2 = 0;        len = strlen(str);        i = 0;        caculate(0,0);        printf("Tree %d: %d => %d\n",t++,height1,height2);    }    return 0;}