[ACM] POJ 2418 Hardwood Species (Trie树或map)

Hardwood Species

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 17986		Accepted: 7138

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.

Source

Waterloo Local 2002.01.26

解题思路：

给出一系列的字符串。有同样的，问每一个字符串出现的次数站总字符串个数的百分比是多少，输出的时候按字符串的字典序输出。

第一种方法：用STL里面的map<string ,int >， first为字符串。int为出现的个数。更重要的是map是默认string按字典序从小到大排列的，所以输出的时候直接从头到尾就能够了。

另外一种方法：建立Trie树，对于输入的每个字符串都插入到树中。每个字符串在树中的最后一个字母节点保存着该字符串出现的次数，并用一个ok标记该节点是不是一个字符串的最后一个字母节点。

注意：折腾了非常长时间，发现用string 保存字符串比用 字符数组慢的太多太多。。。

以后还是用字符数组来保存字符串把。

第一种方法代码：

#include <iostream>
#include <stdio.h>
#include <iomanip>
#include <string.h>
#include <map>
using namespace std;

int main()
{
    map<string,int>mp;
    int cnt=0;
    string s;
    while(getline(cin,s))
    {
        mp[s]++;
        cnt++;
    }
    map<string,int >::iterator i;
    for(i=mp.begin();i!=mp.end();i++)
    {
        cout<<setiosflags(ios::fixed)<<setprecision(4)<<i->first<<" "<<100.0*(i->second)/cnt<<endl;
    }
    return 0;
}

另外一种方法代码：

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <iomanip>
using namespace std;
int num;//字符串的总个数

struct Trie
{
    int cnt;//某个字符串出现的总个数
    char name[40];//保存的字符串
    bool ok;//是不是走到了字符串的最后一个字母，保存最后一个字母的那个节点也保存着整个字符串的Name
    Trie *next[127];//子节点，ASCII最大值为126
    Trie()
    {
        ok=0;
        cnt=0;
        for(int i=0;i<127;i++)
            next[i]=NULL;
    }
}root;

void create(char s[])
{
    int len=strlen(s);
    Trie*p=&root;
    for(int i=0;i<len;i++)
    {
        int id=s[i];
        if(p->next[id]==NULL)
            p->next[id]=new Trie;
        p=p->next[id];
    }
    p->cnt++;//走到字符串的最后一个字母的节点
    strcpy(p->name,s);
    p->ok=1;
}

void dfs(Trie *root)//递归输出
{
    Trie*p=root;
    if(p->ok)
        cout<<p->name<<" "<<setiosflags(ios::fixed)<<setprecision(4)<<100.0*p->cnt/num<<endl;
    for(int i=0;i<127;i++)
    {
        if(p->next[i]!=NULL)
        {
            dfs(p->next[i]);
        }
    }
}

int main()
{
    char s[40];
    while(gets(s))
    {
        create(s);
        num++;
    }
    dfs(&root);
    return 0;
}

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