题目链接：
http://poj.org/problem?id=2418
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Hardwood Species
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 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 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 |
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题目意思：
给n个串，求每个串出现的频率。
n<=1000000
解题思路：
把每个串压到字典树里，维护一个从根节点到当前节点的单词数量。
注意：有空格，大小写字母。空格的ASCII是32，Z的ASCII是90，z的ASCII是122.所以可以都减去32可以转化到0~100进行处理。
代码:
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 110000
struct Node
{
int cnt;
struct Node * next[110]; //空格是32 把空格记为0
}root,node[Maxn*35];
char save[Maxn];
int pp,n;
void inse(char * cur)
{
Node * p=&root;
while(*cur)
{
if(p->next[*cur-‘ ‘]==NULL)
{
node[pp].cnt=0;
memset(node[pp].next,NULL,sizeof(node[pp].next));
p->next[*cur-‘ ‘]=&node[pp++];
}
p=p->next[*cur-‘ ‘];
cur++;
}
p->cnt++;
//printf("%d\n",p->cnt);
}
void cal(Node * cur,string a)
{
//cout<<a<<endl;
//system("pause");
Node * p=cur;
//while(p)
{
if(p->cnt>=1)
{
cout<<a<<‘ ‘;
printf("%.4lf\n",(p->cnt)*100.0/n);
}
for(int i=0;i<100;i++)
{
if(p->next[i])
{
//p=p->next[i];
char temp=i+‘ ‘;
cal(p->next[i],a+temp);
}
}
//p=p->next;
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
n=0,pp=0;
while(gets(save)!=NULL)
{
save[strlen(save)]=‘\0‘;
inse(save);
n++;
}
cal(&root,"");
return 0;
}
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