标签:style blog http color os io for art
poj 2418 Hardwood Species
http://poj.org/problem?id=2418
trie树+dfs
题意: 给你多个单词,问每个单词出现的频率。
方法:通过字典树,将所有单词放入树中,通过dfs遍历(题目要求按ASSIC码顺序输出单词及其频率),dfs可满足
注意:单词中不一定只出现26个英文字母,ASSIC码表共有256个字符
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <cstdio> 6 #include <cstring> 7 #include <cmath> 8 #include <stack> 9 #include <queue> 10 #include <functional> 11 #include <vector> 12 #include <map> 13 using namespace std; 14 #define M 0x0f0f0f0f 15 #define min(a,b) (a>b?b:a) 16 #define max(a,b) (a>b?a:b) 17 int ji; 18 char st[40]; 19 struct tree 20 { 21 int count_; 22 struct tree *next[256]; 23 }; 24 25 void insert_(struct tree *root,char *s) 26 { 27 int i; 28 if(root==NULL||*s==‘\0‘) 29 return ; 30 struct tree *p=root; 31 while(*s!=‘\0‘) 32 { 33 34 if(p->next[*s]==NULL) 35 { 36 struct tree *t=(struct tree *)malloc(sizeof(struct tree)); 37 for(i=0; i<256; i++) 38 t->next[i]=NULL; 39 t->count_=0; 40 p->next[*s]=t; 41 p=t; 42 } 43 else 44 { 45 p=p->next[*s]; 46 } 47 s++; 48 } 49 p->count_++; 50 } 51 int j=0; 52 void dfs(struct tree *p,int j) 53 { 54 int i; 55 if(p->count_) 56 { 57 st[j]=‘\0‘; 58 double m=(p->count_)*1.0/((ji)*1.0)*100; 59 printf("%s %.4lf\n",st,m); 60 } 61 for(i=0; i<256; i++) 62 { 63 if(p->next[i]!=NULL) 64 { 65 st[j]=i; 66 dfs(p->next[i],j+1); 67 } 68 } 69 } 70 71 int main() 72 { 73 char a[10005],a1[10005]; 74 int i; 75 struct tree *root=(struct tree *)malloc(sizeof(struct tree)); 76 for(i=0; i<256; i++) 77 { 78 root->next[i]=NULL; 79 } 80 root->count_=0; 81 ji=0; 82 while(gets(a)!=NULL) 83 { 84 ji++; 85 insert_(root,a); 86 } 87 dfs(root,0); 88 return 0; 89 }
poj 2418 Hardwood Species (trie树),布布扣,bubuko.com
poj 2418 Hardwood Species (trie树)
标签:style blog http color os io for art
原文地址:http://www.cnblogs.com/bibier/p/3886495.html