Huffman和Priority_queue 解决POJ 1521
题目:http://poj.org/problem?id=1521
题目大意:给定字符串,求哈夫曼编码长和它与等长编码的比值
做这道题目的时候wrang了好几次,但是,
经过调试之后,我彻底了解了哈夫曼树的过程
说来相当有价值了。在下面我也会分享出来的。
View Code
#include <iostream> #include "cstdio" #include "string" #include "cstring" #include <queue> using namespace std; struct Num { int number; bool operator<(const Num &a) const { return number>a.number; } }tmp; int main() { string s; char c; int numberofchar, i, j, a, b; while(1) { priority_queue<Num> que; cin>>s; if(s=="END") break; sort(s.begin(), s.end()); c = s[0]; numberofchar = 0; for(i=0; i<s.length(); i++) { if(s[i]==c) numberofchar++; else { tmp.number = numberofchar; que.push(tmp); c = s[i]; numberofchar = 1; } } tmp.number = numberofchar; que.push(tmp); //细节1:这个地方可不能忘了呀,不然,毫无疑问的wrong answer int oldlen = s.length()*8; int newlen = 0; if(que.size()==1) //细节2:这个地方也不能忘了,不然,程序会莫名其妙的出错,自然也是wrong answer newlen = que.top().number; while(que.size()>1) { a = que.top().number; que.pop(); b = que.top().number; que.pop(); tmp.number = a+b; newlen += tmp.number; que.push(tmp); } cout<<oldlen<<" "<<newlen<<" "; printf("%.1f\n", (float)oldlen/newlen); } return 0; }
posted on 2011-09-25 22:06 More study needed. 阅读(313) 评论(0) 编辑 收藏 举报