• PKU2418_树种统计(map应用||Trie树)


    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.

    解法一:map应用(自动按照键值字典序排序,强)

    #include<bits/stdc++.h>
    using namespace std;
    map<string,int>mp;
    int main()
    {
        
        int n;
        cin>>n;
        int t=n;
        getchar();
        string s;
        while(n--){
            getline(cin,s);
            mp[s]++;
        }
        map<string,int>::iterator it;
        for(it=mp.begin();it!=mp.end();it++)
            cout<<it->first<<" "<<setiosflags(ios::fixed)<<setprecision(4)<<100.0*it->second/t<<'%'<<endl;
        return 0;
    }

    解法二:Trie树

    #include<bits/stdc++.h>
    using namespace std;
    int n,len;
    char s[40];
    struct Trie
    {
        int cnt;
        bool ok;
        char name[40];
        Trie *next[127];
        Trie(){
            cnt=0,ok=0;
            memset(next,0,sizeof(next));
        }
    }*root;
    void Insert(char *s,int id)
    {
        Trie *p=root;
        int len=strlen(s);
        for(int i=0;i<len;i++){
            if(!p->next[s[i]-'A'])
                   p->next[s[i]-'A']=new Trie();
            p=p->next[s[i]-'A'];
        }
        p->cnt++; 
        if(!p->ok) p->ok=1;
        strcpy(p->name,s);
    }
    void dfs(Trie *root)
    {
        Trie *p=root;
        if(p->ok){
            printf("%s",p->name); 
            printf(" %.4lf%
    ",100.0*p->cnt/n);
        }
        for(int i=0;i<127;i++){//字典序 
            if(p->next[i])
                dfs(p->next[i]);
        }
    }
    int main()
    {
        cin>>n;
        getchar();
        root=new Trie;
        for(int i=1;i<=n;i++){
            gets(s);
            Insert(s,i);
        }
        dfs(root);
    }
  • 相关阅读:
    现在连Linux都搞不懂,当初我要是这么学习操作系统就好了!
    一时技痒,撸了个动态线程池,源码放Github了
    Java线程池ThreadPoolExecutor使用和分析(一)
    canch----1.对缓存的思考
    1.java 内存数据库--H2数据库使用指南
    What’s the difference between persist, save, merge and update? Which one should you use?
    primary key's generator in JPA entity
    STM32F103驱动M24256 256k存储芯片进行读写
    【Proteus+51单片机学习笔记】-51/52系列单片机简介
    【STM32项目笔记】STM32CubeMX+Keil+Proteus联合实现LED闪烁
  • 原文地址:https://www.cnblogs.com/freinds/p/6744524.html
Copyright © 2020-2023  润新知