标签:style blog http color os for
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1053
讲解:
题意:给定一个字符串,根据哈夫曼编码求出最短长度,并求出比值。
思路:就是哈夫曼编码。把单个字符出现次数作为权值。
AC代码:
1 #include <iostream> 2 #include <string> 3 #include <queue> 4 #include <cstdio> 5 using namespace std; 6 7 class node{ 8 public: 9 int key; //a,b,c... 10 int count;//频率 11 int p;//父亲结点 12 friend bool operator < (const node &a, const node &b) 13 { 14 if(b.count < a.count) 15 return true; 16 else 17 return false; 18 } 19 }; 20 21 int value(char c) 22 { 23 if(c==‘_‘) 24 return 26; 25 else 26 return(c-‘A‘); 27 } 28 int main() 29 { 30 string str; 31 cin >> str; 32 while(str!="END") 33 { 34 node c[60]; 35 for(int i=0;i<60;i++) 36 { 37 c[i].key=i; 38 c[i].count=0; 39 } 40 int length=str.length(); 41 priority_queue<node> q; 42 for(int i=0;i<length;i++) 43 { 44 (c[value(str.at(i))]).count++; 45 } 46 for(int i=0;i<=26;i++) 47 { 48 if(c[i].count!=0) 49 q.push(c[i]); 50 } 51 if(q.size()==1) 52 { 53 printf("%d %d 8.0\n",8*length,length); 54 } 55 else 56 { 57 int high=0; 58 int n=27; 59 while(q.size()>1) 60 { 61 node s1=q.top(); 62 q.pop(); 63 node s2=q.top(); 64 q.pop(); 65 c[n].count=s1.count+s2.count; 66 c[s1.key].p=n; 67 c[s2.key].p=n; 68 high=high+c[n].count; 69 q.push(c[n]); 70 n++; 71 } 72 printf("%d %d %.1lf\n",8*length,high,(((double)(8*length))/((double)high))); 73 } 74 cin >> str; 75 } 76 }
hdoj 1053 Entropy(用哈夫曼编码)优先队列,布布扣,bubuko.com
标签:style blog http color os for
原文地址:http://www.cnblogs.com/lovychen/p/3833221.html