POJ2418 Hardwood Species—二叉查找树应用

1. Hardwood Species原题描述

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 14326		Accepted: 5814

Description

Hardwoods are the botanical group of trees that have broad leaves, produce a fruit or nut, and generally go dormant in the winter.

America's temperate climates produce forests with hundreds of hardwood
species -- trees that share certain biological characteristics. Although
oak, maple and cherry all are types of hardwood trees, for example,
they are different species. Together, all the
hardwood species represent 40 percent of the trees in the United
States.

On the other hand, softwoods, or conifers, from the Latin word meaning
"cone-bearing," have needles. Widely available US softwoods include
cedar, fir, hemlock, pine, redwood, spruce and cypress. In a home, the
softwoods are used primarily as structural lumber
such as 2x4s and 2x6s, with some limited decorative applications.

Using satellite imaging technology, the Department of Natural Resources
has compiled an inventory of every tree standing on a particular day.
You are to compute the total fraction of the tree population represented
by each species.
Input

Input to your program consists of a list
of the species of every tree observed by the satellite; one tree per
line. No species name exceeds 30 characters. There are no more than
10,000 species and no more than 1,000,000 trees.
Output

Print the name of each species represented
in the population, in alphabetical order, followed by the percentage of
the population it represents, to 4 decimal places.
Sample Input

Red Alder
Ash
Aspen
Basswood
Ash
Beech
Yellow Birch
Ash
Cherry
Cottonwood
Ash
Cypress
Red Elm
Gum
Hackberry
White Oak
Hickory
Pecan
Hard Maple
White Oak
Soft Maple
Red Oak
Red Oak
White Oak
Poplan
Sassafras
Sycamore
Black Walnut
Willow

Sample Output

Ash 13.7931
Aspen 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cypress 3.4483
Gum 3.4483
Hackberry 3.4483
Hard Maple 3.4483
Hickory 3.4483
Pecan 3.4483
Poplan 3.4483
Red Alder 3.4483
Red Elm 3.4483
Red Oak 6.8966
Sassafras 3.4483
Soft Maple 3.4483
Sycamore 3.4483
White Oak 10.3448
Willow 3.4483
Yellow Birch 3.4483

2. 理解题意并解答

题目大致意思是输入一堆字符串，可能重复，按字典序输出，并输出占的比例。

当然这道题非常简单，可能大家第一想到的是用hash方法来做，但是hash的方式会造成一定的浪费，且有冲撞的风险，有没有更好的办法呢？当然有，就是使用二叉查找树。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct node
{
char name[31];
struct node *lchild, *rchild;
int count;
}BSTree;

int totalNum;

void mid_visit(BSTree *root, FILE *fp);
void insertBST(BSTree **root, char *str);

int main()
{
BSTree *root = NULL;
char str[31];
FILE *fp;

if((fp = fopen("sample.in", "r")) == NULL)
{
printf("Cannot open file!\n");
exit(1);
}

while (!feof(fp))
{
fgets(str, 30, fp); // 读入一行
if(strlen(str) > 0)  // 防止空行
{
str[strlen(str) - 1] = 0; // 去掉读入的换行符
insertBST(&root, str);
totalNum++;
}
}

fclose(fp);

if((fp = fopen("sample.out", "w")) == NULL)
{
printf("Cannot open file!\n");
exit(1);
}

mid_visit(root, fp);

fclose(fp);

return 0;
}

void mid_visit(BSTree *root, FILE *fp)
{
if(root)
{
mid_visit(root->lchild, fp);
fprintf(fp, "%s %.4lf\n", root->name, (root->count * 100.0) / totalNum);
mid_visit(root->rchild, fp);
}
}

void insertBST(BSTree **root, char *str)
{
if(!*root)
{
BSTree *p = (BSTree*)malloc(sizeof(BSTree));
strcpy(p->name, str); // 注意这里不能直接赋值p->name = str，因为p->name是数组名，不是左值
p->lchild = NULL;
p->rchild = NULL;
p->count = 1;
*root = p;
}
else
{
if(strcmp((*root)->name, str) == 0)
{
(*root)->count++;
return;
}
else if(strcmp((*root)->name, str) > 0) // 注意*root必须加()，因为->优先级大于*
{
insertBST(&(*root)->lchild, str);
}
else
{
insertBST(&(*root)->rchild, str);
}
}
}

可能有人会想，干嘛怎么麻烦？C++中不是有map吗？map就是红黑树实现的，红黑树是一种高效的二叉查找树，它通过红黑结点的约束，保证了树的高度总是在2logn以下，性能比二叉查找树更高。是的，用C++的话就方便了许多。

#include <iostream>
#include <fstream>
#include <string>
#include <map>
#include <iomanip>

using namespace std;

int main()
{
string str;
int totalNum = 0;
map<string, int> tree;

ifstream fin("sample.in");
ofstream fout("sample.out");

while(getline(fin, str))
{
tree[str]++;
totalNum++;
}

map<string, int>::iterator itr;
for(itr = tree.begin(); itr != tree.end(); itr++)
{
fout << itr->first << ' '<< fixed << setprecision(4)
<< itr->second * 100.0 / totalNum << endl;
}

fin.close();
fout.close();

return 0;
}

发现代码量减少了一半，说明C++的STL库还是非常值得熟悉和掌握的，特别是在参加一些比赛时，能节省不少时间。如果是在平时练习，那么不用STL也是不错的选择，因为可以锻炼我们深入理解二叉查找树，其它也类似。