为了好玩和荣耀,这里是一个基于 Boost Spirit 的实现。我添加了更多的虚假投票输入,以便可以显示一些东西。
我不确定候选人和选票之间是否存在 1:1 的关系(我不是美国公民,我不知道列出的候选人是否会投票或被投票)。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";
然而,该代码可用于这两种目的。
- 我让输入出现的顺序变得不重要。
- 但是,您可以选择断言单个字母 (S,M,J) 实际上对应于该点之前列出的位置。通过取消注释检查启用此功能
posletter_check
在http://liveworkspace.org/code/d9e39c19674fbf7b2419ff88a642dc38上查看演示
#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, phx::ref(*this), _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 *****