POJ 2418, ZOJ 1899 Hardwood Species-from lanshui_Yang

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 2x4 s and 2x6 s, with some limited decorative applications.

Using satellite imaging technology, the Department of Natural Resources has compiled an inventory of every tree standing on a particle 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 1, 13.7931
Aspen 1, 3.4483
Basswood 3.4483
Beech 3.4483
Black Walnut 3.4483
Cherry 3.4483
Cottonwood 3.4483
Cyber press 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
Here are some strings for you to calculate the probability of each string.

Solution: I wrote this question using the Trie tree. map may use TLE. The output string must be in the Lexicographic Order, so I used dfs.

Ps: This question is a single set of data on POJ, but multiple groups of data on ZOJ. Please note that! In addition, the strings in this question may contain other characters except letters. Be careful!

See the Code:

 

# Include <iostream> # include <cstring> # include <cstdio> # include <algorithm> # define mem (a, B) memset (a, B, sizeof ()) using namespace std; const int MAXN = 3e5 + 5; char A [50]; int vis [MAXN]; // create A tag array for the final output string int cnt; // set the number int sum for each node in the Trie tree; // count the number of strings. struct Tnode {int count; // count the number of words displayed at the node. char f; // record the character int xu stored in the node in the Trie tree; // record the node number in the Trie tree Tnode * next [256]; // increase the array size, because it may contain other characters except letters. Tnode () {count = 0; f = 0; xu = 0; mem (next, 0) ;}; char s [50]; void inse (char * str, tnode * root) {Tnode * p = root; int id; while (* str) {id = * str-'\ 0 '; if (p-> next [id] = 0) {p-> next [id] = new Tnode; p-> next [id]-> xu = ++ cnt; p-> next [id]-> f = * str;} p = p-> next [id]; ++ str;} p-> count ++ ;} double ans; void print (int deep, Tnode * p) // dfs outputs the string in Lexicographic Order, deep For recursive depth {int I; for (I = 0; I <256; I ++) {if (p-> next [I]! = NULL) {int tmp = p-> next [I]-> xu; A [deep] = p-> next [I]-> f; if (p-> next [I]-> count> 0 &&! Vis [tmp]) {vis [tmp] = 1; ans = p-> next [I]-> count * 100.0/sum *; int j; for (j = 0; j <= deep; j ++) {printf ("% c", A [j]);} printf ("%. 4f \ n ", ans);} print (deep + 1, p-> next [I]) ;}} int main () {bool flag = false; while (gets (s) {if (flag) // an empty row printf ("\ n") is output between two adjacent groups of data; else flag = true; cnt = 0; sum = 0; Tnode * root = new Tnode; memset (vis, 0, sizeof (vis); inse (s, root); sum ++; while (gets (s) & strlen (s) {inse (s, root); sum ++;} print (0, root);} return 0 ;}