POJ 2418 Hardwood Species (字典树||MAP)
Hardwood Species
| Time Limit: 10000MS | Memory Limit: 65536K | |
| Total Submissions: 20684 | Accepted: 8108 |
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.
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 3.4483Yellow Birch 3.4483
题目大意：就是查看所有的树种在在全部树中所占的比例。
思路：裸的字典树或這MAP，看完题目后，1)如何建树有些茫然，因为之前建的树都是全为大写或小写字母，这个是什么都有。。
2)如何按照字典序输出，且不重复。
Trie:
#include<iostream> #include<cstring>//用模板加string #include<cstdio> #include<queue> #include<algorithm> #define SUM 250001 #define LL long long #define inf 0x3f3f3f3f using namespace std; int s; struct node { int vis; node *next[301]; }; struct no { char s[32]; int r; }w[10010]; char now[32]; int sum=0,cnt=0; node *init() { node *p; p=new node ; memset(p->next,NULL,sizeof(p->next));//因为不知道每个字符的范围，所以干脆全部都治空 p->vis=0;//用于记录当前单词出现的次数 return p; } int op(no a,no b) { return strcmp(a.s,b.s)<0; } void bu(node *root,char *now) { char *c=now;//<span id="transmark"></span>将当前的单词，赋值给C，便于将此单词保存在w[].s中 while(*now!='\0') { int i=*now;//直接将此单词的ASCII作为下标进行建树 if(root->next[i]==NULL) { root->next[i]=init(); } root=root->next[i]; now++;//移动下一个字节 } if(root->vis==0) { strcpy(w[sum++].s,c);//注意 赋值的时候不能将now赋值，因为此时的now已经是单词的结尾， } root->vis++; } void look(node *root,char *per) { char *c=per; int i,j,k; while(*per!='\0') { i=*per; if(!root->next[i]) return ; root=root->next[i]; per++; } printf("%s %.4f\n",c,(float)(root->vis*100)/cnt);//如果没有在执行到当前语句返回，证明则树中有此单词 } int main() { int i,j,k; node *root; root=new node ; root=init();//初始化 while(gets(now)!=NULL) { if(!strcmp(now,"")) break; cnt++;//记录总共的树的数量 bu(root,now);//建树 } sort(w,w+sum,op);//进行W内的字符串排序 for(i=0;i<sum;i++) { look(root,w[i].s); } return 0; }map:
#include <iostream> #include <cstdio> #include <cstring> #include<string> #include <map> #include<algorithm> #include<stdlib.h> #include<ctype.h> #include<queue> #include<math.h> char s[40]; using namespace std; int main() { int n,m,i; double x; map<string,int>mp; map<string,int>::iterator cnt; n=0; while(gets(s)!=NULL) { if(!strcmp(s,""))//有无此句都对只不过在Linux下不能直接出结果，若无需要CTRL+D break; n++; mp[s]++; } for(cnt=mp.begin(); cnt!=mp.end(); cnt++) { x=(cnt->second*100.0)/n; printf("%s %.4lf\n",cnt->first.c_str(),x); } return 0; }