Huffman和Priority_queue 解決POJ 1521
題目:http://poj.org/problem?id=1521
題目大意:給定字符串,求哈夫曼編碼長(zhǎng)和它與等長(zhǎng)編碼的比值
做這道題目的時(shí)候wrang了好幾次,但是,
經(jīng)過(guò)調(diào)試之后,我徹底了解了哈夫曼樹的過(guò)程
說(shuō)來(lái)相當(dāng)有價(jià)值了。在下面我也會(huì)分享出來(lái)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); //細(xì)節(jié)1:這個(gè)地方可不能忘了呀,不然,毫無(wú)疑問(wèn)的wrong answer int oldlen = s.length()*8; int newlen = 0; if(que.size()==1) //細(xì)節(jié)2:這個(gè)地方也不能忘了,不然,程序會(huì)莫名其妙的出錯(cuò),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. 閱讀(327) 評(píng)論(0) 收藏 舉報(bào)

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