ACM 粗心永远AC不了系列——UVa 213 Message Decoding(World Finals1991,字符串) |二维数组的妙用
2017-03-13 13:55
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此题在《算法竞赛入门经典》第二版中 第83页 里面有详细的讲解;,本人仅再此写下读书的理解心得其中二维数组的应用思想很巧妙,这里着重提出一下,审题后发现一个特点,
相同的二进制数对应的编码字符由二进制长度不同而不同,如‘001’与‘01’对应的就是两个不同的编码
由两种因素决定值,就可以用二维数组表示,当然这里用一维数组表示也可以,只不过会做一个由长度和二进制值而映射到单个数字的映射函数,不够直观。
解中code[len][v]存储字符,其中len表示编码长度,v表示编码该长度下二进制对应的编码编号,这两个因素就可以存储题中编码。
书上核心代码
#include <stdio.h>
#include <string.h>
#pragma warning(disable:4996)
#define maxcnt 10000
char a[maxcnt];
int readchar();
int readint(int c);
int code[8][1 << 8];//1<<8==2^8,存放编码头;len为相应二进制数转换为的十进制,1<<8意思是ch的范围就是256,所以数组开那么大就好了
int readcodes();
int main(){
while (readcodes()){
while (1){
int len = readint(3);//取得编码长度
if (len == 0)break;//为'000'则退出
while (1){
int v = readint(len);//取得一个单位的编码,由二进制转化为十进制
if (v == (1 << len) - 1)break;//1<<len==2^len 这里的意思是你的编码全为1的话,如v='1111' (len=4)则退出
putchar(code[len][v]);
}
}
putchar('\n');
}
return 0;
}
//读取下一个非换行符字符,以int形式输出
int readchar(){
while (1){
int ch = getchar();
if (ch != '\n'&&ch != '\r')return ch;
}
}
//将若干位的二进制数转换为10进制数
int readint(int c){
int v = 0;
while (c--)v = v * 2 + readchar() - '0';
return v;
}
//将编码赋值到数组中,分别是 code[编码长度][先后顺序]=编码字符
int readcodes(){
memset(code, 0, sizeof(code));
code[1][0] = readchar();
for (int len = 2; len <= 7; len++){
for (int i = 0; i < (1 << len) - 1; i++){
int ch = getchar();
if (ch == EOF)return 0;
if (ch == '\r' || ch == '\n')return 1;
code[len][i] = ch;
}
}
}
相同的二进制数对应的编码字符由二进制长度不同而不同,如‘001’与‘01’对应的就是两个不同的编码
由两种因素决定值,就可以用二维数组表示,当然这里用一维数组表示也可以,只不过会做一个由长度和二进制值而映射到单个数字的映射函数,不够直观。
解中code[len][v]存储字符,其中len表示编码长度,v表示编码该长度下二进制对应的编码编号,这两个因素就可以存储题中编码。
Message Decoding
Some message encoding schemes require that an encoded message be sent in two parts. The first part, called the header, contains the characters of the message. The second part contains a pattern that represents the message. You must write a program that can decode messages under such a scheme.The heart of the encoding scheme for your program is a sequence of ``key" strings of 0's and 1's as follows:The first key in the sequence is of length 1, the next 3 are of length 2, the next 7 of length 3, the next 15 of length 4, etc. If two adjacent keys have the same length, the second can be obtained from the first by adding 1 (base 2). Notice that there are no keys in the sequence that consist only of 1's.The keys are mapped to the characters in the header in order. That is, the first key (0) is mapped to the first character in the header, the second key (00) to the second character in the header, the kth key is mapped to the kth character in the header. For example, suppose the header is:AB#TANCnrtXcThen 0 is mapped to A, 00 to B, 01 to #, 10 to T, 000 to A, ..., 110 to X, and 0000 to c.The encoded message contains only 0's and 1's and possibly carriage returns, which are to be ignored. The message is divided into segments. The first 3 digits of a segment give the binary representation of the length of the keys in the segment. For example, if the first 3 digits are 010, then the remainder of the segment consists of keys of length 2 (00, 01, or 10). The end of the segment is a string of 1's which is the same length as the length of the keys in the segment. So a segment of keys of length 2 is terminated by 11. The entire encoded message is terminated by 000 (which would signify a segment in which the keys have length 0). The message is decoded by translating the keys in the segments one-at-a-time into the header characters to which they have been mapped.Input
The input file contains several data sets. Each data set consists of a header, which is on a single line by itself, and a message, which may extend over several lines. The length of the header is limited only by the fact that key strings have a maximum length of 7 (111 in binary). If there are multiple copies of a character in a header, then several keys will map to that character. The encoded message contains only 0's and 1's, and it is a legitimate encoding according to the described scheme. That is, the message segments begin with the 3-digit length sequence and end with the appropriate sequence of 1's. The keys in any given segment are all of the same length, and they all correspond to characters in the header. The message is terminated by 000.Carriage returns may appear anywhere within the message part. They are not to be considered as part of the message.Output
For each data set, your program must write its decoded message on a separate line. There should not be blank lines between messages.Sample input
TNM AEIOU 0010101100011 1010001001110110011 11000 $#**\ 0100000101101100011100101000
Sample output
TAN ME ##*\$
书上核心代码
#include <stdio.h>
#include <string.h>
#pragma warning(disable:4996)
#define maxcnt 10000
char a[maxcnt];
int readchar();
int readint(int c);
int code[8][1 << 8];//1<<8==2^8,存放编码头;len为相应二进制数转换为的十进制,1<<8意思是ch的范围就是256,所以数组开那么大就好了
int readcodes();
int main(){
while (readcodes()){
while (1){
int len = readint(3);//取得编码长度
if (len == 0)break;//为'000'则退出
while (1){
int v = readint(len);//取得一个单位的编码,由二进制转化为十进制
if (v == (1 << len) - 1)break;//1<<len==2^len 这里的意思是你的编码全为1的话,如v='1111' (len=4)则退出
putchar(code[len][v]);
}
}
putchar('\n');
}
return 0;
}
//读取下一个非换行符字符,以int形式输出
int readchar(){
while (1){
int ch = getchar();
if (ch != '\n'&&ch != '\r')return ch;
}
}
//将若干位的二进制数转换为10进制数
int readint(int c){
int v = 0;
while (c--)v = v * 2 + readchar() - '0';
return v;
}
//将编码赋值到数组中,分别是 code[编码长度][先后顺序]=编码字符
int readcodes(){
memset(code, 0, sizeof(code));
code[1][0] = readchar();
for (int len = 2; len <= 7; len++){
for (int i = 0; i < (1 << len) - 1; i++){
int ch = getchar();
if (ch == EOF)return 0;
if (ch == '\r' || ch == '\n')return 1;
code[len][i] = ch;
}
}
}
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