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poj 2418 -- Hardwood Species

2024-07-14 16:23:52   223人阅读

Hardwood Species
Time Limit: 10000MS Memory Limit: 65536K
Total Submissions: 18174 Accepted: 7206

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.
 
思路：统计出现的百分比。字典树水过。
 
 1 /*====================================================================== 2  *           Author :   kevin 3  *         Filename :   HardwoodSpecies.cpp 4  *       Creat time :   2014-07-29 10:53 5  *      Description : 6 ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio>10 #include <cstring>11 #include <queue>12 #include <cmath>13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 12815 using namespace std;16 struct Trie{17     Trie *next[M];18     int cnt;19     //int v;20 };21 Trie *root;22 int num,z;23 char ss[35];24 void CreateTrie(char *str)25 {26     int len = strlen(str);27     Trie *p= root,*q;28     for(int i = 0; i < len; i++){29         int id = str[i];30         if(p->next[id] == NULL){31             q = (Trie *)malloc(sizeof(Trie));32             //q->v = 1;33             for(int j = 0; j < M; j++){34                 q->next[j] = NULL;35                 q->cnt = 0;36             }37             p->next[id] = q;38             p = p->next[id];39         }40         else{41             //p->next[id]->v++;42             p = p->next[id];43         }44     }45     //p->v = -1;46     p->cnt++;47 }48 void Count(Trie * tr)49 {50     if(tr->cnt){51         printf("%s %.4lf\n",ss,100*((double)tr->cnt / (double)num));52     }53     for(int i = 0; i < M; i++){54         if(tr->next[i]){55             ss[z++] = i;56             ss[z] = ‘\0‘;57             Count(tr->next[i]);58             z--;59         }60     }61 }62 int DealTrie(Trie *T)63 {64     if(T == NULL) return 0;65     for(int i = 0; i < M; i++){66         if(T->next[i])67             DealTrie(T->next[i]);68     }69     free(T);70     return 0;71 }72 int main(int argc,char *argv[])73 {74     root = (Trie*)malloc(sizeof(Trie));75     for(int i = 0; i < M; i++){76         root->next[i] = NULL;77         root->cnt = 0;78     }79     char str[35];80     num = 0;81     z = 0;82     while(gets(str) != NULL){83         num++;84         CreateTrie(str);85     }86     Count(root);87     DealTrie(root);88     return 0;89 }
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