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.
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
开始想用map水一下。。。 结果超时。。。
但是C++能过 ，G++就过不了
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii pair<string, int>
using namespace std;
char ch[50];
int main ()
{
map <string,int >mp;
map<string,int>::iterator it;
int coun=0;
while(gets(ch)&&strlen(ch))
{
string str(ch);
it = mp.find(str);
if( it != mp.end())
it->second = it->second+1;
else
mp.insert(pii(str,1));
coun++;
}
for( it=mp.begin() ; it != mp.end(); ++it){
cout << it->first;
printf(" %.4f
",(it->second)*1.0/coun*100);
}
return 0;
}
#include <iostream>
#include <stdio.h>
#include <string.h>
#define MAX 128
using namespace std;
char A[50];
int coun = 0;
struct Trie
{
int endmark;
Trie*next[128];
Trie()
{
endmark=0;
for(int i=0;i<128;i++)
next[i]=NULL;
}
}root;
void create(char *s)//将字符串s建立在trie树中
{
Trie *p=&root;
int len=strlen(s);
for(int i=0;i<len;i++)
{
int id=s[i];//唯一标识
if(p->next[id]==NULL)
p->next[id]=new Trie;
p=p->next[id];
}
p->endmark++;
}
void print(int deep , Trie * p) // dfs 按字典序输出字符串 ， deep为递归的深度
{
if(p->endmark > 0 ){
for(int j=0;j<deep;j++)
printf("%c",A[j]);
printf(" %.4f
",p->endmark*1.0/coun*100);
return ;
}
for(int i = 0 ; i < 128; i ++ )
{
if(p -> next[i] != NULL){
A[deep] = char(i) ;
//cout<<"---"<<(char)i<<" "<<p->next[i]->endmark<<"----
";
print(deep+1 , p -> next[i]) ;
}
}
return ;
}
int main()
{
char s[35];
while(gets(s)&&strlen(s)!=0){
create(s);//根据给定的字符串建立字典树
coun++;
}
print(0,&root);
return 0;
}