poj 2418 Hardwood Species

Hardwood Species

Time Limit: 10000MS

Memory Limit: 65536K

Total Submissions: 20931

Accepted: 8208

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

题目链接

解题思路： 本题有两种解法，分别使用map和字典序做本题。

法一：map

使用map让本题中的每一种树对应一个int值，在输入过程中，如果前面没有这种树，就创建一个新值，即map[str] = 1,如果有这种树，就在原有基础上加1，即map[str]++;输入结束后使用sort对其进行排序，再依次输出。这种方法简单，运行内存低，但时间复杂度高，达到3141ms.

代码如下：

#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <map>

using namespace std;

const int maxn = 1000010;

string str,name[10010];
int ans=0,tot=0;
map<string,int> tree;

int main()
{
while(getline(cin,str) && str!="")
{
if(tree.find(str) == tree.end())
{
tree[str] = 1;
name[ans++] = str;
}
else
tree[str]++;
tot++;
}
sort(name,name+ans);
for(int i=0;i<ans;i++)
{
printf("%s %.4lf\n",name[i].c_str(),tree[name[i]]*1.0/tot*100);
}
return 0;
}

法二：字典序

本题也可用字典序外加深搜解决，首先在输入过程中建树，输入结束后，在使用dfs按照字典序对其进行深搜，终止条件是如果p->cnt不等于0就输出。

代码如下：

<span style="font-size:14px;">#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

struct node{    //建立节点
int cnt;
node *next[130];
node()
{
cnt = 0;
memset(next,0,sizeof(next));
}
};

int n;
node *root = NULL;

void buildtrie(char *s)//建树
{
node *tmp = NULL;
node *p = root;
int l = strlen(s);
for(int i=0;i<l;i++)
{
if(p->next[s[i]] == NULL)
{
tmp = new node;
p->next[s[i]] = tmp;
}
p = p->next[s[i]];
}
p->cnt++;//每个单词末尾cnt都不为0
}

void dfs(node *root)
{
static char word[35];
static int pos;
int i;
if(root->cnt)//终止条件，如果找到尾巴就输出
{
word[pos] = '\0';
printf("%s %.4lf\n",word,root->cnt*100.0/n);
return;
}
for(int i=0;i<130;i++)//按照字典序从头开始遍历
{
if(root->next[i])
{
word[pos++] = i;
dfs(root->next[i]);
pos--;
}
}
}

int main()
{
char str[35];
n = 0;
root = new node;
while(gets(str) && strcmp(str,"") != 0)
{
buildtrie(str);
n++;
}
dfs(root);
return 0;
}
</span>