POJ 2418 Hardwood Species

POJ 2418 Hardwood Species

题目传送门

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceeded.

题目翻译：

给定若干字符串，输出格式为：（按字典序）给出的字符串+这个字符串出现次数在给定字符串总数之中的占比。

题解：

介绍三种东西：

第一种——字符串忽略空格只考虑回车的输入方式。

第二种——(C++\,\,STL\,\,map)容器。

第三种——字典树。

第一种：

字符串忽略空格只考虑回车的输入方式。

代码如下：

getline(cin,s);

（具体的使用方法可以去网上看，但是讲的都过于繁琐，各种流什么玩意的专业术语根本看不懂（可能是蒟蒻太弱了）），我个人认为大家只需要记住这个玩意的格式就可以。

这里的s指代的是(C++STL)中的(string)容器。如果有对这个容器还不是很了解的小伙伴，请移步本蒟蒻的这篇博客：

浅谈C++ STL string容器

第二种：

map容器。

如果有对map容器不是很熟悉的小伙伴请参考本蒟蒻的这篇博客：

详解C++ STL map容器

第三种：

字典树。

如果有对字典树还不是很了解的小伙伴请参考本蒟蒻的这篇博客：

详解字典树

---

好了，我们现在来说思路：

题目让我们统计字符串出现的次数，那么我们很容易想到(map)所构建的这种映射关系，建立一个(map)构建由(string)到(int)的一个映射就可以方便的统计这些信息。

最令人舒适的是，(map)自带的第一关键字排序正好支持(string)的字典序操作。简直不要太舒服！

于是得出了一份超短AC代码：

#include<cstdio>
#include<map>
#include<iostream>
#include<string>
using namespace std;
int tot;
string s;
map<string,int> mp;
map<string,int>::iterator it;
int main()
{
    while(getline(cin,s))
    {
        tot++;
        mp[s]++;
    }
    for(it=mp.begin();it!=mp.end();it++)
    {
        double ans=(it->second)*100.0/tot;
        printf("%s %.4lf
",it->first.c_str(),ans);
    }
    return 0;
}