Hardwood Species——物种所占比例

Hardwood Species

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

题意：在一个森林里，有很多物种，现已知所有的树木，问每种树占所有树的比例是多少。

解析：问每种树木所占的比例是多少，首先要知道每种树的数目以及所有树的总数目。所有树的总数目很容易知道，输入了多少次就有多少树，但是要把不同的树木进行分类，然后统计没种树的数目。这里用了两种方法，一种是字典树，一种是MAP，这里先讲字典树。

将每种树木都写进字典树中，树的种类用结构体数组单独存储，每种树的数目通过在字典树中添加的时候确定，如果已经有这种树了，那么返回这种树的存储地址，使地址加一；如果没有这种树，返回新增物种，数目初始化为1。具体看代码：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define MAX 1<<16

using namespace std;

struct Trie
{
    int next[127];
    int sign;
} trie[900000];
int top = 0,n = 0;
struct Tree
{
    char stree[30];
    int sum;
} tree[10009];

int cmp(Tree a,Tree b)
{
    if(strcmp(a.stree,b.stree) <= 0)
        return 1;
    else
        return 0;
}

int newTrie()
{
    memset(trie[top].next,-1,sizeof(trie[top].next));
    trie[top].sign = -1;
    return top++;
}
int init(int k,char *sz)
{
    int l,j,i;
    l = strlen(sz);
    for(i = 0; i < l; i++)
    {
        j = sz[i];
        if(trie[k].next[j] == -1)
            trie[k].next[j] = newTrie();
        k = trie[k].next[j];
    }
    if(trie[k].sign == -1)
    {
        trie[k].sign = n;
        return -1;
    }
    return trie[k].sign;
}

int main()
{
    char ss[31];
    int k,j,i,num = 0;
    top = 0;
    k = newTrie();
    n = 0;
    while(gets(ss))
    {
        num++;
        int flag = 0;
        flag = init(k,ss);
        if(flag == -1)
        {
            strcpy(tree[n].stree,ss);
            tree[n].sum = 1;
            n++;
        }
        else
        {
            tree[flag].sum++;
        }
    }
    sort(tree,tree+n,cmp);
    for(i = 0; i < n; i++)
    {
        double jie = (double)tree[i].sum/(double)num;
        printf("%s %.4lf\n",tree[i].stree,jie*100);
    }
    return 0;
}

MAP实现

map看起来是很简单的，而且也很好理解，简单说来就是将逗号后面的映射到前面去，前面的就相当于数组的下标，而后面的就相当于数组内存储的内容。默认字典序排列

这里用到了迭代器，是一个指针变量，用来遍历map容器并输出。

#include <iostream>  
#include <cstdio>  
#include <string>  
#include <cstring>  
#include <stdlib.h>  
#include <math.h>  
#include <ctype.h>  
#include <queue>  
#include <map>  
#include<algorithm>  
  
using namespace std;  

int main()
{
    map<string,int>q;
    map<string,int>::iterator i;
    int num = 0;
    char sz[40];
    while(gets(sz))
    {
        if(sz[0] == '\0')
            break;
        q[sz]++;
        num++;
    }
    for(i = q.begin();i != q.end();i++)
    {
        double x = (i->second * 100.0) / num;
        printf("%s %.4lf\n",i->first.c_str(),x);
    }
    return 0;
}