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            poj2418

            Hardwood Species

            Time Limit: 10000MS Memory Limit: 65536K
            Total Submissions: 13263 Accepted: 5434

            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 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 3.4483
            Yellow 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

            所有字符都有
            水的trie樹+dfs
            #include <cstdio>
            #include 
            <cstdlib>
            #include 
            <cstring>
            #include 
            <cmath>
            #include 
            <ctime>
            #include 
            <cassert>
            #include 
            <iostream>
            #include 
            <sstream>
            #include 
            <fstream>
            #include 
            <map>
            #include 
            <set>
            #include 
            <vector>
            #include 
            <queue>
            #include 
            <algorithm>
            #include 
            <iomanip>
            #define maxn 40000
            using namespace std;
            struct node
            {
                
            int next[256];
                
            int count;
                
            void init()
                
            {
                    memset(next,
            -1,sizeof(next));
                    count
            =0;
                }

            }
            s[maxn];
            int sind;
            int num,ans,n;
            char str[50];
            void cas_init()
            {
                s[
            0].init();
                sind
            =1;
            }

            int ins(char str[])
            {
                
            int i,j,ind;
                ind
            =0;
                
            for(int i=0;i<strlen(str);i++)
                
            {
                    j
            =str[i];
                    
            if(s[ind].next[j]==-1)
                    
            {
                        s[sind].init();
                        s[ind].next[j]
            =sind++;
                    }

                    ind
            =s[ind].next[j];
                }

                
            if(s[ind].count!=0)
                
            {
                    s[ind].count
            ++;
                    
            return 0;
                }

                
            else
                
            {
                    s[ind].count
            ++;
                    
            return 1;
                }

            }

            void dfs(int id,int depth)
            {
                
            if(s[id].count>0)
                
            {
                    str[depth]
            ='\0';
                    printf(
            "%s %.4lf\n",str,(s[id].count*100.0/double(num)));
                }

                
            for(int i=0;i<256;i++)
                
            if(s[id].next[i]!=-1)
                
            {
                    str[depth]
            =i;
                    dfs(s[id].next[i],depth
            +1);
                }

            }

            int main()
            {
                
            int flag;
                
            char str[40];
                n
            =0;num=0;
                
            //printf("%d",strcmp("!","@"));
                
            //freopen("in1.txt","r+",stdin);
                cas_init();
                
            while(gets(str))
                
            {
                    num
            ++;
                    flag
            =ins(str);
                    
            if(flag==1) n++;
                }

                dfs(
            0,0);
                
            //fclose(stdin);
                return 0;
            }


            posted on 2012-07-24 13:53 jh818012 閱讀(176) 評論(0)  編輯 收藏 引用


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            • 1.?re: poj1426
            • 我嚓,,輝哥,,居然搜到你的題解了
            • --season
            • 2.?re: poj3083
            • @王私江
              (8+i)&3 相當于是 取余3的意思 因為 3 的 二進制是 000011 和(8+i)
            • --游客
            • 3.?re: poj3414[未登錄]
            • @王私江
              0ms
            • --jh818012
            • 4.?re: poj3414
            • 200+行,跑了多少ms呢?我的130+行哦,你菜啦,哈哈。
            • --王私江
            • 5.?re: poj1426
            • 評論內(nèi)容較長,點擊標題查看
            • --王私江
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