hdu 1053 Entropy
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=1053
Entropy
Description
An entropy encoder is a data encoding method that achieves lossless data compression by encoding a message with “wasted” or “extra” information removed. In other words, entropy encoding removes information that was not necessary in the first place to accurately encode the message. A high degree of entropy implies a message with a great deal of wasted information; english text encoded in ASCII is an example of a message type that has very high entropy. Already compressed messages, such as JPEG graphics or ZIP archives, have very little entropy and do not benefit from further attempts at entropy encoding.
English text encoded in ASCII has a high degree of entropy because all characters are encoded using the same number of bits, eight. It is a known fact that the letters E, L, N, R, S and T occur at a considerably higher frequency than do most other letters in english text. If a way could be found to encode just these letters with four bits, then the new encoding would be smaller, would contain all the original information, and would have less entropy. ASCII uses a fixed number of bits for a reason, however: it’s easy, since one is always dealing with a fixed number of bits to represent each possible glyph or character. How would an encoding scheme that used four bits for the above letters be able to distinguish between the four-bit codes and eight-bit codes? This seemingly difficult problem is solved using what is known as a “prefix-free variable-length” encoding.
In such an encoding, any number of bits can be used to represent any glyph, and glyphs not present in the message are simply not encoded. However, in order to be able to recover the information, no bit pattern that encodes a glyph is allowed to be the prefix of any other encoding bit pattern. This allows the encoded bitstream to be read bit by bit, and whenever a set of bits is encountered that represents a glyph, that glyph can be decoded. If the prefix-free constraint was not enforced, then such a decoding would be impossible.
Consider the text “AAAAABCD”. Using ASCII, encoding this would require 64 bits. If, instead, we encode “A” with the bit pattern “00”, “B” with “01”, “C” with “10”, and “D” with “11” then we can encode this text in only 16 bits; the resulting bit pattern would be “0000000000011011”. This is still a fixed-length encoding, however; we’re using two bits per glyph instead of eight. Since the glyph “A” occurs with greater frequency, could we do better by encoding it with fewer bits? In fact we can, but in order to maintain a prefix-free encoding, some of the other bit patterns will become longer than two bits. An optimal encoding is to encode “A” with “0”, “B” with “10”, “C” with “110”, and “D” with “111”. (This is clearly not the only optimal encoding, as it is obvious that the encodings for B, C and D could be interchanged freely for any given encoding without increasing the size of the final encoded message.) Using this encoding, the message encodes in only 13 bits to “0000010110111”, a compression ratio of 4.9 to 1 (that is, each bit in the final encoded message represents as much information as did 4.9 bits in the original encoding). Read through this bit pattern from left to right and you’ll see that the prefix-free encoding makes it simple to decode this into the original text even though the codes have varying bit lengths.
As a second example, consider the text “THE CAT IN THE HAT”. In this text, the letter “T” and the space character both occur with the highest frequency, so they will clearly have the shortest encoding bit patterns in an optimal encoding. The letters “C”, “I’ and “N” only occur once, however, so they will have the longest codes.
There are many possible sets of prefix-free variable-length bit patterns that would yield the optimal encoding, that is, that would allow the text to be encoded in the fewest number of bits. One such optimal encoding is to encode spaces with “00”, “A” with “100”, “C” with “1110”, “E” with “1111”, “H” with “110”, “I” with “1010”, “N” with “1011” and “T” with “01”. The optimal encoding therefore requires only 51 bits compared to the 144 that would be necessary to encode the message with 8-bit ASCII encoding, a compression ratio of 2.8 to 1.
Input
The input file will contain a list of text strings, one per line. The text strings will consist only of uppercase alphanumeric characters and underscores (which are used in place of spaces). The end of the input will be signalled by a line containing only the word “END” as the text string. This line should not be processed.
Output
For each text string in the input, output the length in bits of the 8-bit ASCII encoding, the length in bits of an optimal prefix-free variable-length encoding, and the compression ratio accurate to one decimal point.
Sample Input
AAAAABCD
THE_CAT_IN_THE_HAT
END
Sample Output
64 13 4.9
144 51 2.8
哈弗曼编码。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<queue>
#include<set>
using std::set;
using std::sort;
using std::swap;
using std::string;
using std::priority_queue;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 30;
const int INF = 0x3f3f3f3f;
typedef unsigned long long ull;
char buf[11000];
struct Node {
char dat;
int w;
Node *ch[2];
Node(char _dat_, int _w_, Node *l = NULL, Node *r = NULL) {
dat = _dat_, w = _w_;
ch[0] = l, ch[1] = r;
}
Node(const Node &x) {
dat = x.dat, w = x.w;
ch[0] = x.ch[0], ch[1] = x.ch[1];
}
inline bool operator<(const Node &x) const {
return w > x.w;
}
};
struct Hoffmancode {
int sum;
Node *root;
priority_queue<Node> q;
inline char to_lower(char ch) {
return ch >= 'A' && ch <= 'Z' ? ch + 32 : ch;
}
inline void CreateHoffmanTree() {
Node *l = NULL, *r = NULL;
while (!q.empty() && q.size() != 1) {
l = new Node(q.top()); q.pop();
r = new Node(q.top()); q.pop();
Node ret(0, l->w + r->w, l, r);
q.push(ret);
}
if (!q.empty()) {
root = new Node(q.top()); q.pop();
}
}
inline void CreateHoffmanCode(Node *x, string str) {
if (!x) return;
if (x->dat != 0) {
sum += (str.length() * x->w);
}
CreateHoffmanCode(x->ch[0], str + "0");
CreateHoffmanCode(x->ch[1], str + "1");
}
inline void solve() {
sum = 0, root = NULL;
int arr[N] = { 0 }, tot = 0, n = strlen(buf);
while (!q.empty()) q.pop();
for (int i = 0; i < n; i++) {
arr[to_lower(buf[i]) - '_']++;
}
for (int i = 0; i < 28; i++) {
if (arr[i]) {
tot++;
Node ret('_' + i, arr[i]);
q.push(ret);
}
}
if (1 == tot) {
printf("%d %d 8.0\n", n * 8, n);
return;
}
CreateHoffmanTree();
CreateHoffmanCode(root, "");
printf("%d %d %.1lf\n", n * 8, sum, (double)(n * 8) / sum);
}
}work;
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
while (~scanf("%s", buf) && 0 != strcmp(buf, "END")) {
work.solve();
}
return 0;
}
hdu 1053 Entropy的更多相关文章
- HDU 1053 Entropy(哈夫曼编码 贪心+优先队列)
传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1053 Entropy Time Limit: 2000/1000 MS (Java/Others) ...
- hdu 1053 Entropy (哈夫曼树)
Entropy Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Sub ...
- HDU 1053 & HDU 2527 哈夫曼编码
http://acm.hdu.edu.cn/showproblem.php?pid=1053 #include <iostream> #include <cstdio> #in ...
- hdoj 1053 Entropy(用哈夫曼编码)优先队列
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 ...
- hdu 1053 (huffman coding, greedy algorithm, std::partition, std::priority_queue ) 分类: hdoj 2015-06-18 19:11 22人阅读 评论(0) 收藏
huffman coding, greedy algorithm. std::priority_queue, std::partition, when i use the three commente ...
- 【HDOJ】1053 Entropy
构造huffman编码,果断对字符进行状态压缩. #include <iostream> #include <cstdio> #include <cstring> ...
- HDU题解索引
HDU 1000 A + B Problem I/O HDU 1001 Sum Problem 数学 HDU 1002 A + B Problem II 高精度加法 HDU 1003 Maxsu ...
- CSU-ACM2018暑期训练7-贪心
A:合并果子(贪心+优先队列) B:HDU 1789 Doing Homework again(非常经典的贪心) C:11572 - Unique Snowflakes(贪心,两指针滑动保存子段最大长 ...
- hdu 2527 Safe Or Unsafe (哈夫曼树)
Safe Or Unsafe Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)To ...
随机推荐
- vb 取得桌面路径
txtPath.Text = System.Environment.GetFolderPath(Environment.SpecialFolder.Desktop)
- 启动obiee
启动obiee:1.启动数据库第一步:打开Oracle监听$ lsnrctl start第二步:使用sysdab角色登录sqlplussqlplus / as sysdba第三步:启动数据库SQL&g ...
- C语言中的fread和fwrite
C语言中的fread和fwrite是专门用来操作文件的方法. 1. fread负责从打开的文件指针中读取文件内容. 函数原型:size_t fread(void *p, size_t size, si ...
- webview渲染流程
文档标记说明 ################# 消息边界 +++++++++++++++++ 区域分隔 $$$$$$$$$$$$$$$$$ 线程边界 ~~~~~~~~~~~~~~~~~ 进程边界 - ...
- 《Linux企业应用案例精解(第2版)》新书开始发售
<Linux企业应用案例精解(第2版)>新书开始发售 650) this.width=650;" title="linux企业应用案例精解 第2版" alt= ...
- js的二元三元操作符
二元 if ( a == b) { alert(a) } // (a == b) && alert(a) if ( a != b) { alert(a) } // (a == b) | ...
- 【Python】django模型models的外键关联使用
Python 2.7.10,django 1.8.6 外键关联:http://www.bubuko.com/infodetail-618303.html 字段属性:http://www.cnblogs ...
- 模仿QQ空间 网页设计
目的:1.通过模仿QQ空间,全自主写代码,熟悉网页设计的流程 2.熟练的掌握HTML.CSS.JS的应用 3.将在此过程中遇到的问题及其解决方法记录在此,以便取用. 开始: 一.登陆界面(index. ...
- javascript 回车提交指定按钮
当页面上有多个提交按钮时,使用回车键会触发第一个按钮的点击事件.现在我们想触发指定按钮的提交,只需要在最后输入的文本框中加入 onkeydown 事件,如下 <asp:TextBox ID=&q ...
- .NET中的字符串你了解多少?
字符串的特性 1.不可变性 由于字符串是不可变的的,每次修改字符串,都是创建了一个单独字符串副本(拷贝了一个字符串副本).之所以发生改变只是因为指向了一块新的地址. ps: ...