UVa 1593 Alignment of Code
2015-03-15 16:24
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Alignment of Code
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Description
![](http://uva.onlinejudge.org/components/com_onlinejudge/images/button_pdf.png)
You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is basically a text editor with bells and whistles. You are working on a module that takes a piece of code containing
some definitions or other tabular information and aligns each column on a fixed vertical position, while keeping the resulting code as short as possible, making sure that only whitespaces that are absolutely required stay in the code. So, that the first words
on each line are printed at position p1 = 1; the second words on each line are printed at the minimal possible position p2, such that all first words end at or before
position p2 - 2; the third words on each line are printed at the minimal possible position p3, such that all second words end at or before position p3 -
2, etc.
For the purpose of this problem, the code consists of multiple lines. Each line consists of one or more words separated by spaces. Each word can contain uppercase and lowercase Latin letters, all ASCII punctuation
marks, separators, and other non-whitespace ASCII characters (ASCII codes 33 to 126 inclusive). Whitespace consists of space characters (ASCII code 32).
word, each word is 1 to 80 characters long (inclusive). Words are separated by one or more spaces. Lines of the code can have both leading and trailing spaces. Each line in the input file is at most 180 characters long. There are at most 1000 lines in the
input file.
that i-th word on each line starts at the same position pi.
Note for the Sample:
The `
![](http://uva.onlinejudge.org/external/15/4983img1.png)
' character in the example below denotes a space
character in the actual files (ASCII code 32).
Sample
Sample
分析:
因为每一行不知道有多少字符串,故考虑用vector<>。
读入每一行可用fgets(),也可以是getline(),再用stringstream读入vector。
每一列输出长度由该列最长的单词决定,所以先遍历每一列,保存求出的最大长度。
最后是输出格式,行末决不能有多余空格,即当某行某一列是最后一列时,不再输出空格。
代码:
Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
![](http://uva.onlinejudge.org/components/com_onlinejudge/images/button_pdf.png)
You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is basically a text editor with bells and whistles. You are working on a module that takes a piece of code containing
some definitions or other tabular information and aligns each column on a fixed vertical position, while keeping the resulting code as short as possible, making sure that only whitespaces that are absolutely required stay in the code. So, that the first words
on each line are printed at position p1 = 1; the second words on each line are printed at the minimal possible position p2, such that all first words end at or before
position p2 - 2; the third words on each line are printed at the minimal possible position p3, such that all second words end at or before position p3 -
2, etc.
For the purpose of this problem, the code consists of multiple lines. Each line consists of one or more words separated by spaces. Each word can contain uppercase and lowercase Latin letters, all ASCII punctuation
marks, separators, and other non-whitespace ASCII characters (ASCII codes 33 to 126 inclusive). Whitespace consists of space characters (ASCII code 32).
Input
The input file contains one or more lines of the code up to the end of file. All lines (including the last one) are terminated by a standard end-of-line sequence in the file. Each line contains at least oneword, each word is 1 to 80 characters long (inclusive). Words are separated by one or more spaces. Lines of the code can have both leading and trailing spaces. Each line in the input file is at most 180 characters long. There are at most 1000 lines in the
input file.
Output
Write to the output file the reformatted, aligned code that consists of the same number of lines, with the same words in the same order, without trailing and leading spaces, separated by one or more spaces suchthat i-th word on each line starts at the same position pi.
Note for the Sample:
The `
![](http://uva.onlinejudge.org/external/15/4983img1.png)
' character in the example below denotes a space
character in the actual files (ASCII code 32).
Sample
Input
start: integer; // begins here stop: integer; // ends here s: string; c: char; // temp
Sample
Output
start: integer; // begins here stop: integer; // ends here s: string; c: char; // temp
分析:
因为每一行不知道有多少字符串,故考虑用vector<>。
读入每一行可用fgets(),也可以是getline(),再用stringstream读入vector。
每一列输出长度由该列最长的单词决定,所以先遍历每一列,保存求出的最大长度。
最后是输出格式,行末决不能有多余空格,即当某行某一列是最后一列时,不再输出空格。
代码:
#include <iostream> #include <sstream> #include <string> #include <vector> #include <cstdio> #define LOCAL using namespace std; const int maxn = 1010; vector<string> str[maxn]; int len[maxn]; int main() { #ifndef LOCAL freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); #endif int n = 0, max_col = 0; string s, buf; while (getline(cin, s)) { stringstream ss(s); while (ss >> buf) str .push_back(buf); if (str .size() > max_col) max_col = str .size(); n++; } for (int i = 0; i < max_col; i++) { int mlen = 0; for (int j = 0; j < n; j++) { if (str[j].size() >= i + 1) { if (str[j][i].length() > mlen) mlen = str[j][i].length(); } len[i] = mlen; } } for (int i = 0; i < n; i++) { for (int j = 0; j < max_col; j++) { if (str[i].size() >= j + 1) { cout << str[i][j]; if (str[i].size() > j + 1) for (int k = str[i][j].length(); k < len[j]; k++) cout << " "; } if (j < str[i].size() - 1) cout << " "; } cout << "\n"; } return 0; }
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