POJ 2418 简单trie树
Hardwood Species
Time Limit: 10000MS Memory Limit: 65536K
Total Submissions: 21845 Accepted: 8551
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.
10000ms的time limit……….
题意:每行一个字符串,统计不同的字符串出现的百分比。最后按ASCII排序输出不同字符串和出现的百分比。
思路:
1.很好理解的简单trie树(如果我这句话伤害到其他人了,我收回这句话,。,。,。)
2.简单快排(这个真真简单,30行基本就搞定了)
3.STL神器:map
4.看到Discuss里面有用BST这个神奇的东西的(表示不会)
5.一个hash
这道题有一些比较坑的地方,建议G++WA的同学们交C++!!!
我同样的程序G++WA了,C++AC 鬼知道为什么。
另外,输入要用gets()。gets()是到’\n’结束的,空格不是结束的标志,正好适用此题。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct trie
{
int num;
trie *next[256];
};
trie *root=new trie;
char ans[55];
int cnt=0;
void insert(char ch[])
{
trie *p=root,*newtrie;
for(int i=0;ch[i]!='\0';i++)
{
if(p->next[ch[i]]==NULL)
{
newtrie=new trie;
for(int j=0;j<256;j++)
{
newtrie->next[j]=NULL;
}
newtrie->num=0;
p->next[ch[i]]=newtrie;
p=newtrie;
}
else
{
p=p->next[ch[i]];
}
}
p->num++;
}
void dfs(trie *p,int t)
{
if(p->num!=0)
{
ans[t]=0;
printf("%s %.4lf\n",ans,100*p->num/(1.0*cnt));
}
for(int i=0;i<256;i++)
{
if(p->next[i]!=NULL)
{
ans[t]=i;
dfs(p->next[i],t+1);
}
}
}
int main()
{
for(int i=0;i<256;i++) root->next[i]=NULL;
root->num=0;
char s[45];
while(gets(s))
{
insert(s);
cnt++;
}
dfs(root,0);
}
POJ 2418 简单trie树的更多相关文章
- POJ3630——简单Trie树
这个题的意思是说,给出一些字符串,判断是否有字符串是另一个字符串的前缀,当然可以用排序水过,不过这个题拿来练习一下Trie树不错. 这个题在poj的discuss上好多人说必须要静态建树,估计都是用了 ...
- Colored Sticks POJ - 2513(trie树欧拉路)
题意: 就是无向图欧拉路 解析: 不能用map..超时 在判断是否只有一个联通的时候,我比较喜欢用set,但也不能用set,会超时,反正不能用stl emm 用trie树来编号就好了 #include ...
- POJ 3764 DFS+trie树
题意: 给你一棵树,求树中最长的xor路径.(n<=100000) 思路: 首先我们知道 A xor B =(A xor C) xor (B xor C) 我们可以随便选一个点DFS 顺便做出与 ...
- trie树--详解
文章作者:yx_th000 文章来源:Cherish_yimi (http://www.cnblogs.com/cherish_yimi/) 转载请注明,谢谢合作.关键词:trie trie树 数据结 ...
- 转:trie树--详解
前几天学习了并查集和trie树,这里总结一下trie. 本文讨论一棵最简单的trie树,基于英文26个字母组成的字符串,讨论插入字符串.判断前缀是否存在.查找字符串等基本操作:至于trie树的删除单个 ...
- 电话号码 【trie树】
电话号码 查看 提交 统计 提问 总时间限制: 1000ms 内存限制: 65536kB 描写叙述 给你一些电话号码,请推断它们是否是一致的,即是否有某个电话是还有一个电话的前缀. 比方: Emerg ...
- POJ 3630 Phone List(trie树的简单应用)
题目链接:http://poj.org/problem?id=3630 题意:给你多个字符串,如果其中任意两个字符串满足一个是另一个的前缀,那么输出NO,否则输出YES 思路:简单的trie树应用,插 ...
- [ACM] POJ 2418 Hardwood Species (Trie树或map)
Hardwood Species Time Limit: 10000MS Memory Limit: 65536K Total Submissions: 17986 Accepted: 713 ...
- POJ 2418 Hardwood Species 【Trie树】
<题目链接> 题目大意: 给你一堆字符串,让你按字典序输出他们出现的频率. 解题分析: 首先,这是Trie数词频统计的题目,以Trie树的边储存字母,节点存储以该节点结尾的链所代表的字符串 ...
随机推荐
- 8 switch case
当一个case成立,从这个case向后穿透所有case,即使后面的case条件不成立 包括default,直到程序结束或者遇到break程序才结束. 1.case是常量,且不能重复 2.表达式可以是b ...
- PAT-day1
1001 害死人不偿命的(3n+1)猜想 (15 分) 卡拉兹(Callatz)猜想: 对任何一个正整数 n,如果它是偶数,那么把它砍掉一半:如果它是奇数,那么把 ( 3n+1)砍掉一半.这样一直 ...
- PAT_A1018#Public Bike Management
Source: PAT A1018 Public Bike Management (30 分) Description: There is a public bike service in Hangz ...
- react 父组件给子组件传值
父组件 import React from 'react'import '../page1/header.css'import { Table } from 'antd'import Child fr ...
- NOIP2018 滚粗记
Day -2 上午,大家都在复习各种模板,zhx总结了足足67个模板(杨辉三角也算模板???),lgl死磕FFT发现cos和sin打反了,我也是复习板子和以前做过的题,几乎没有人颓. 接着jdr,l ...
- C#学习笔记_07_数组
07_数组 数组的声明与实例化 名词解释 数组:数组是一个容器,用来存储一系列相兼容的数据类型的变量: 实例化:声明一个数组,并且赋初始值: 数组长度:就是数组的容量,表示这个数组可以存储多少个数据: ...
- hdu2016 数据的交换输出【C++】
数据的交换输出 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Sub ...
- 【codeforces 776E】The Holmes Children
[题目链接]:http://codeforces.com/contest/776/problem/E [题意] f(n)是小于n的不同整数对(x,y)这里x+y==n且gcd(x,y)==1的个数; ...
- 【LeetCode Weekly Contest 26 Q4】Split Array with Equal Sum
[题目链接]:https://leetcode.com/contest/leetcode-weekly-contest-26/problems/split-array-with-equal-sum/ ...
- poj 3621最优比例生成环(01分数规划问题)
/* 和求最小生成树差不多 转载思路:http://www.cnblogs.com/wally/p/3228171.html 思路:之前做过最小比率生成树,也是属于0/1整数划分问题,这次碰到这道最优 ...