
Republican Senator John McMahon Democrat Mayor Steven Markel Republican Judge Matt Stevens Democrat Senator Anthony Quizitano S R M D J R ... .. ..
格式基本上是前3行包括Party,position和name,接下来的行表示“结果”,其形式为:
[第一封信] [党投票]
所以,例如,如果你看到S R,这意味着对参议员投了1票,那就是共和党候选人.
这是我到目前为止所拥有的:
#include<fstream> int main() { std::ifstream input("file.txt"); } 这是我的理解,这将允许我输入文件,并逐行进行,但我不知道如何从这里开始实现这个…任何帮助?
谢谢!
解决方法 为了娱乐和荣耀,这是一个基于Boost Spirit的实现.我添加了更多假投票输入,只是因此可能会显示一些内容.>我不确定候选人和投票之间是否存在1:1的关系(我不是美国公民,我不知道列出的候选人是投票还是投票).所以我决定只使用假数据.
const std::string input = "Republican Senator John McMahon\n" "Democrat Senator Anthony Quizitano\n" "S R\n" "S R\n" "S R\n" "Democrat Mayor Steven Markel\n" "Republican Judge Matt Stevens\n" "M D\n" "J R\n" "S R\n" "S R\n";
但是,该代码可用于这两个目的.
>我将输入显示的顺序变得不重要.
>但是,您可以断言,单个字母(S,M,J)实际上对应于该点之前列出的位置.通过使用posletter_check取消注释检查来启用此功能
请参阅http://liveworkspace.org/code/d9e39c19674fbf7b2419ff88a642dc38的现场演示
#define BOOST_SPIRIT_USE_PHOENIX_V3#define BOOST_RESulT_OF_USE_DECLTYPE#include <boost/fusion/adapted.hpp>#include <boost/spirit/include/qi.hpp>#include <boost/spirit/include/phoenix.hpp>#include <iomanip>namespace qi = boost::spirit::qi;namespace phx = boost::phoenix;struct CandIDate { std::string party,position,name; };BOOST_FUSION_ADAPT_STRUCT(CandIDate,(std::string,party)(std::string,position)(std::string,name))typedef std::map<std::pair<char,char>,size_t> Votes;typedef std::vector<CandIDate> CandIDates;template <typename It> struct parser : qi::grammar<It>{ mutable Votes _Votes; mutable CandIDates _candIDates; parser() : parser::base_type(start) { using namespace qi; using phx::bind; using phx::ref; using phx::val; start = (line % eol) >> *eol >> eoi; line = Vote [ phx::bind(&parser::register_Vote,phx::ref(*this),_1) ] | candIDate [ phx::push_back(phx::ref(_candIDates),_1) ] ; Vote %= graph // Comment the following line to accept any single // letter,even if a matching position wasn't seen // before: [ _pass = phx::bind(&parser::posletter_check,_1) ] >> ' ' >> char_("RD") ; candIDate = (string("Republican") | string("Democrat")) >> ' ' >> as_string [ +graph ] >> ' ' >> as_string [ +(char_ - eol) ] ; } private: bool posletter_check(char posletter) const { for (auto& c : _candIDates) if (posletter == c.position[0]) return true; return false; } voID register_Vote(Votes::key_type const& key) const { auto it = _Votes.find(key); if (_Votes.end()==it) _Votes[key] = 1; else it->second++; } qi::rule<It,Votes::key_type()> Vote; qi::rule<It,CandIDate()> candIDate; qi::rule<It> start,line;};int main(){ const std::string input = "Republican Senator John McMahon\n" "Democrat Senator Anthony Quizitano\n" "S R\n" "S R\n" "S R\n" "Democrat Mayor Steven Markel\n" "Republican Judge Matt Stevens\n" "M D\n" "J R\n" "S R\n" "S R\n"; std::string::const_iterator f(std::begin(input)),l(std::end(input)); parser<std::string::const_iterator> p; try { bool ok = qi::parse(f,l,p); if (ok) { std::cout << "\ncandIDate List\n"; std::cout << "------------------------------------------------\n"; for (auto& c : p._candIDates) std::cout << std::setw(20) << c.name << " (" << c.position << " for the " << c.party << "s)\n"; std::cout << "\nVote distribution:\n"; std::cout << "------------------------------------------------\n"; for (auto& v : p._Votes) std::cout << '(' << v.first.first << "," << v.first.second << "): " << v.second << " Votes " << std::string(v.second,'*') << "\n"; } else std::cerr << "parse Failed: '" << std::string(f,l) << "'\n"; if (f!=l) std::cerr << "trailing unparsed: '" << std::string(f,l) << "'\n"; } catch(const qi::expectation_failure<std::string::const_iterator>& e) { std::string frag(e.first,e.last); std::cerr << e.what() << "'" << frag << "'\n"; }} 输出:
candIDate List------------------------------------------------ John McMahon (Senator for the Republicans) Anthony Quizitano (Senator for the Democrats) Steven Markel (Mayor for the Democrats) Matt Stevens (Judge for the Republicans)Vote distribution:------------------------------------------------(J,R): 1 Votes *(M,D): 1 Votes *(S,R): 5 Votes *****总结
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