HDU 3753 Alignment of Code 字符串对其 stl
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Alignment of Code
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1132 Accepted Submission(s): 270
Problem Description
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).
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 begins with an integer T. The next T blocks each represents a case. Each case contains one or more lines of the code up to the end of case mark-a single '@' in a single line. All lines are terminated by a standard end-of-line sequence in the file. Each line contains at least one 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 for each case.
Output
For each case, 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 such that i-th word on each line starts at the same position pi.
Sample Input
1 start: integer; // begins herestop: integer; // ends here s: string; c: char; // temp@
Sample Output
start: integer; // begins herestop: integer; // ends heres: string;c: char; // temp
Author
Roman Elizarov
Source
2010 Northeastern European Regional Contest
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题目大意:让一组字符串的每一列的单词左对齐。
分析:大体分为四步:
1.将每一句中的单词拆分开,获得单个的单词
2.获取每一列的最大宽度,也就是最长的单词的长度
3.获取每一行的单词的个数
4.输出时,按照每一列的宽度设置单词的宽度,注意最后面不能有空格
因此,问题的核心就是如何将每一个单词拆分开,当然可以通过扫描并以空格为分隔符进行手工拆分,但STL还提供了更加方便的函数:istringstream(string str),头文件为<sstream>。控制宽度时自然想到setw(),因为要左对其,还要使用setiosflags(ios::left).
#include <iostream>#include <sstream>#include <memory.h>#include <iomanip>using namespace std;string line;//保存一行数据string block[1010][190];int word_len[190];//每一列的单词最长长度int word_num[1010];//每一行的单词数量int r,c;int t;void output(){ int i,j; cout<<setiosflags(ios::left);//左对齐 for(i=0;i<r;i++) { for(j=0;j<word_num[i]-1;j++) { cout<<setw(word_len[j]+1)<<block[i][j];//控制宽度 } cout<<block[i][j]<<endl;//最后一行不加空格,单独输出 }}void input(){ string temp; r=c=0; memset(word_len,0,sizeof(word_len)); memset(word_num,0,sizeof(word_num)); while(getline(cin,temp)&&temp!="@") { if(temp=="") continue; istringstream cut_str(temp); while(cut_str>>temp) { block[r][c]=temp; word_len[c]=max((int)temp.size(),word_len[c]); c++; } word_num[r]=c; c=0; r++; }}int main(){ cin>>t; while(t--) { input(); output(); } return 0;}
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