[字典树] poj 2418 Hardwood Species

题目链接：

http://poj.org/problem?id=2418

Hardwood Species

Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 17511 Accepted: 6949

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 AlderAshAspenBasswoodAshBeechYellow BirchAshCherryCottonwoodAshCypressRed ElmGumHackberryWhite OakHickoryPecanHard MapleWhite OakSoft MapleRed OakRed OakWhite OakPoplanSassafrasSycamoreBlack WalnutWillow

Sample Output

Ash 13.7931Aspen 3.4483Basswood 3.4483Beech 3.4483Black Walnut 3.4483Cherry 3.4483Cottonwood 3.4483Cypress 3.4483Gum 3.4483Hackberry 3.4483Hard Maple 3.4483Hickory 3.4483Pecan 3.4483Poplan 3.4483Red Alder 3.4483Red Elm 3.4483Red Oak 6.8966Sassafras 3.4483Soft Maple 3.4483Sycamore 3.4483White Oak 10.3448Willow 3.4483Yellow Birch 3.4483

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceeded.

Source

Waterloo Local 2002.01.26

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题目意思：

给n个串，求每个串出现的频率。

n<=1000000

解题思路：

把每个串压到字典树里，维护一个从根节点到当前节点的单词数量。

注意：有空格，大小写字母。空格的ASCII是32，Z的ASCII是90，z的ASCII是122.所以可以都减去32可以转化到0~100进行处理。

代码:

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 110000struct Node{    int cnt;    struct Node * next[110]; //空格是32 把空格记为0}root,node[Maxn*35];char save[Maxn];int pp,n;void inse(char * cur){    Node * p=&root;    while(*cur)    {        if(p->next[*cur-' ']==NULL)        {            node[pp].cnt=0;            memset(node[pp].next,NULL,sizeof(node[pp].next));            p->next[*cur-' ']=&node[pp++];        }        p=p->next[*cur-' '];        cur++;    }    p->cnt++;    //printf("%d\n",p->cnt);}void cal(Node * cur,string a){    //cout<<a<<endl;    //system("pause");    Node * p=cur;    //while(p)    {        if(p->cnt>=1)        {            cout<<a<<' ';            printf("%.4lf\n",(p->cnt)*100.0/n);        }        for(int i=0;i<100;i++)        {            if(p->next[i])            {                //p=p->next[i];                char temp=i+' ';                cal(p->next[i],a+temp);            }        }        //p=p->next;    }}int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   n=0,pp=0;   while(gets(save)!=NULL)   {       save[strlen(save)]='\0';       inse(save);       n++;   }   cal(&root,"");   return 0;}