Morris' Blog 我在摸索我存在的意、生命的意、涵及值。 有真正神手般的害，套用模式是大部分的情。 那害的我存在的意何，托出神的存在？ 有候不能怪人不我，我已始省思， 什我有人的原因，而是全要靠人。 有候我我不同，那是因我做不到很多人能到的事情。 我不到、玩不起、得不明、理解得不多， 我缺少的文能力，就是一切一切能的判定。 此，我不得能做些人有助的事情。 我知道我自己是不服的性格，但在我不得不。 我有能力，因此我必。 我到底世界作什？充不定性， 而我是法出拔萃，想一般人似乎也回不了了。 越大的挑需要越力的支持，而我的支持是什？ 最近人，但都是失的局，人能接受， 不需要言，不需要通，一切所想都表不出。

49的鼓 6站台

Problem C: Hardwood 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.

The first line is the number of test cases, followed by a blank line. Each test case of 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. There is a blank line between each consecutive test case.

For each test case 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. Print a blank line between 2 consecutive data sets.

Sample Input

1 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 

Output for Sample Input

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

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Trie {
 int n;
 int link[128];
}Node[400000];
int TrieSize;
void insertTrie(const char* str) {
 static int i, idx;
 idx = 0;
 for(i = 0; str[i]; i++) {
 if(Node[idx].link[str[i]] == 0) {
 TrieSize++;
 memset(&Node[TrieSize], 0, sizeof(Node[0]));
 Node[idx].link[str[i]] = TrieSize;
 }
 idx = Node[idx].link[str[i]];
 }
 Node[idx].n ++;
}
char buf[35];
void printAns(int n, int now, int slen) {
 if(Node[now].n != 0) {
 buf[slen] = '\0';
 printf("%s %.4f\n", buf, Node[now].n*100.0/n);
 }
 int i;
 for(i = 0; i < 128; i++) {
 if(Node[now].link[i] != 0) {
 buf[slen] = i;
 printAns(n, Node[now].link[i], slen+1);
 }
 }
}
int main() {
 int t;
 char buf[35];
 scanf("%d", &t);
 getchar();
 getchar();
 while(t--) {
 int n = 0;
 memset(&Node[0], 0, sizeof(Node[0]));
 TrieSize = 0;
 while(gets(buf)) {
 if(buf[0] == '\0') 
 break;
 insertTrie(buf);
 n++;
 }
 printAns(n, 0, 0);
 if(t)
 puts("");
 }
 return 0;
}
 

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