4

我正在寻找一个从双精度映射到对象指针的容器。但是，每个键只是对应于该对象的一个​​双精度范围。

例如，可能存在 <(0.0 3.0), ptr> 或 <(3.5 10.0), ptr2> 的键/值对

container[1.0] 应该返回 ptr，container[3.0] 也应该返回 ptr，并且 container[-1.0] 应该是未定义的。

默认情况下是否有任何具有类似行为的对象，或者我必须自己实现它？

编辑

这是我编写的实际代码，可能更容易调试/提供建议。

// Behavior: A range is defined mathematically as (min, max]

class dblRange
{
public:
    double min;
    double max;

    dblRange(double min, double max)
    {
        this->min = min;
        this->max = max;
    };

    dblRange(double val)
    {
        this->min = val;
        this->max = val;
    };

    int compare(const dblRange rhs)
    {
        // 1 if this > rhs
        // 0 if this == rhs
        //-1 if this < rhs
        if (rhs.min == rhs.max && min == max)
        {
            /*if (min > rhs.min)
                return 1;
            else if (min == rhs.min)
                return 0;
            else
                return -1;*/
            throw "You should not be comparing values like this. :(\n";
        }
        else if (rhs.max == rhs.min)
        {
            if (min > rhs.min) 
                return 1;
            else if (min <= rhs.min && max > rhs.min)
                return 0;
            else // (max <= rhs.min)
                return -1;
        }
        else if (min == max)
        {
            if (min >= rhs.max)
                return 1;
            else if (min < rhs.max && min >= rhs.min)
                return 0;
            else // if (min < rhs.min
                return -1;
        }

        // Check if the two ranges are equal:
        if (rhs.min == min && rhs.max == max)
        {
            return 0;
        }
        else if (rhs.min < min && rhs.max <= min)
        {
            // This is what happens if rhs is fully lower than this one.
            return 1;
        }
        else if (rhs.min > min && rhs.min >= max)
        {
            return -1;
        }
        else
        {
            // This means there's an undefined case. Ranges are overlapping, 
            // so comparisons don't work quite nicely.

            throw "Ranges are overlapping weirdly. :(\n";
        }
    };

    int compare(const dblRange rhs) const
    {
        // 1 if this > rhs
        // 0 if this == rhs
        //-1 if this < rhs
        if (rhs.min == rhs.max && min == max)
        {
            /*if (min > rhs.min)
                return 1;
            else if (min == rhs.min)
                return 0;
            else
                return -1;*/
            throw "You should not be comparing values like this. :(\n";
        }
        else if (rhs.max == rhs.min)
        {
            if (min > rhs.min) 
                return 1;
            else if (min <= rhs.min && max > rhs.min)
                return 0;
            else // (max <= rhs.min)
                return -1;
        }
        else if (min == max)
        {
            if (min >= rhs.max)
                return 1;
            else if (min < rhs.max && min >= rhs.min)
                return 0;
            else // if (min < rhs.min
                return -1;
        }

        // Check if the two ranges are equal:
        if (rhs.min == min && rhs.max == max)
        {
            return 0;
        }
        else if (rhs.min < min && rhs.max <= min)
        {
            // This is what happens if rhs is fully lower than this one.
            return 1;
        }
        else if (rhs.min > min && rhs.min >= max)
        {
            return -1;
        }
        else
        {
            // This means there's an undefined case. Ranges are overlapping, 
            // so comparisons don't work quite nicely.

            throw "Ranges are overlapping weirdly. :(\n";
        }
    };

    bool operator== (const dblRange rhs ) {return (*this).compare(rhs)==0;};
    bool operator== (const dblRange rhs ) const {return (*this).compare(rhs)==0;};
    bool operator!= (const dblRange rhs ) {return (*this).compare(rhs)!=0;};
    bool operator!= (const dblRange rhs ) const {return (*this).compare(rhs)!=0;};
    bool operator< (const dblRange rhs ) {return (*this).compare(rhs)<0;};
    bool operator< (const dblRange rhs ) const {return (*this).compare(rhs)<0;};
    bool operator> (const dblRange rhs ) {return (*this).compare(rhs)>0;};
    bool operator> (const dblRange rhs ) const {return (*this).compare(rhs)>0;};
    bool operator<= (const dblRange rhs ) {return (*this).compare(rhs)<=0;};
    bool operator<= (const dblRange rhs ) const {return (*this).compare(rhs)<=0;};
    bool operator>= (const dblRange rhs ) {return (*this).compare(rhs)>=0;};
    bool operator>= (const dblRange rhs ) const {return (*this).compare(rhs)>=0;};

};

现在，即使定义了比较运算符，我也无法让地图接受双精度作为键。

这是我用来测试它是否可以工作的一些驱动代码：

std::map<dblRange, int> map;
map[dblRange(0,1)] = 1;
map[dblRange(1,4)] = 2;
map[dblRange(4,5)] = 3;

map[3.0] = 4;
4

7 回答 7

14

I mostly agree with Earwicker in that you can define a range. Now, I am in favor of implementing operators with the real meaning (do what basic types do: two ranges compare equal if both ranges ARE equal). Then you can use the third map parameter to pass it a comparison functor (or function) that solves your particular problem with this map.

// Generic range, can be parametrized for any type (double, float, int...)
template< typename T >
class range
{
public:
    typedef T value_type;

    range( T const & center ) : min_( center ), max_( center ) {}
    range( T const & min, T const & max )
        : min_( min ), max_( max ) {}
    T min() const { return min_; }
    T max() const { return max_; }
private:
    T min_;
    T max_;
};

// Detection of outside of range to the left (smaller values):
//
// a range lhs is left (smaller) of another range if both lhs.min() and lhs.max() 
// are smaller than rhs.min().
template <typename T>
struct left_of_range : public std::binary_function< range<T>, range<T>, bool >
{
    bool operator()( range<T> const & lhs, range<T> const & rhs ) const
    {
        return lhs.min() < rhs.min()
            && lhs.max() <= rhs.min();
    }
};
int main()
{
    typedef std::map< range<double>, std::string, left_of_range<double> > map_type;

    map_type integer; // integer part of a decimal number:

    integer[ range<double>( 0.0, 1.0 ) ] = "zero";
    integer[ range<double>( 1.0, 2.0 ) ] = "one";
    integer[ range<double>( 2.0, 3.0 ) ] = "two";
    // ...

    std::cout << integer[ range<double>( 0.5 ) ] << std::endl; // zero
    std::cout << integer[ range<double>( 1.0 ) ] << std::endl; // one
    std::cout << integer[ 1.5 ] << std::endl; // one, again, implicit conversion kicks in
}

You must be careful with equality and comparisons among double values. Different ways of getting to the same value (in the real world) can yield slightly different floating point results.

于 2009-07-06T22:48:29.293 回答
6

Create a class DoubleRange to store the double range, and implement the comparison operators on it. That way, std::map will do the rest for you, with the DoubleRange class as the key.

于 2009-07-06T21:01:55.670 回答
2

It is better to use Interval tree data structure. Boost has an implementation in Interval Container Library

于 2013-12-06T13:35:05.407 回答
0

One approach would be to calculate the "break points" before hand:

typedef vector< tuple<double, double, foo*> > collisionlist_t;
const collisionlist_t vec;
vec.push_back(make_tuple(0.0, 3.0, ptr));
vec.push_back(make_tuple(3.5, 10.0, ptr2));
// sort 
std::map<double, foo*> range_lower_bounds;
for(collisionlist_t::const_iterator curr(vec.begin()), end(vec.end()); curr!=end; ++curr)
{
    /* if ranges are potentially overlapping, put some code here to handle it */
    range_lower_bounds[curr->get<0>()] = curr->get<2>();
    range_lower_bounds[curr->get<1>()] = NULL;
}

double x = // ...
std::map<double, foo*>::const_iterator citer = range_lower_bounds.lower_bound(x);
于 2009-07-06T22:17:38.903 回答
0

Another suggestion: Use a mathematical transform to map the index from REAL to INT which can be directly compared.

If these ranges are multiple and dense there's also a structure known as an "interval tree" which may help.

于 2009-07-08T05:35:30.973 回答
0

Are the intervals open or closed or half open? I will assumed closed. Note that the intervals cannot overlap by the definition of a map. You will also need splitting rules for when one inserts an over lapping interval. the rules need to decide where the split takes place and must take into account floating point epsilon.

this implementation uses map::lower_bound and does NOT use a class as the domain of the map

map::lower_bound returns an iterator to the first element in a map with a key value that is equal to or greater than that of a specified key. (ie the least key greater than or equal to K. An unfortunate choice of STL method names as it is the least upper bound of K.)

template <class codomain>
class RangeMap : private std::map<double,std::pair<double,codomain>{
public:
    typedef double domain;
    typedef std::map<double,std::pair<double,codomain>:: super;
    typename super::value_type value_type;
protected:
    static domain& lower(const value_type& v){
        return v.first;
    }

    static domain& upper(const value_type& v){
        return v.second.first;
    }

    static codomain& v(const value_type& v){
        return v.second.second;
    }

public:

    static const domain& lower(const value_type& v){
        return v.first;
    }
    static const domain& upper(const value_type& v){
        return v.second.first;
    }
    static const codomain& v(const value_type& v){
        return v.second.second;
    }


    static bool is_point(const value_type& vf) {
        return lower(v) == upper(v);
    }

    static bool is_in(const domain& d,const value_type& vf) {
        return (lower(v) <= d) && (d <= upper(v));
    }


    const_iterator greatest_lower_bound(const domain& d)const {
        const_iterator j = super::lower_bound(d);
        if(j!=end() && j->first==d) return j;//d is the lh side of the closed interval
                                             //remember j->first >= d because it was lower but its the first
        if(j==begin()) return end();//d < all intervals
        --j;                        //back up
        return j;
    }
    const_iterator find(domain& d) {
        const_iterator j = greatest_lower_bound(d);
        if (is_in(j,d)) return j;
        return end();
    }
    iterator greatest_lower_bound(const domain& d) {
        iterator j = super::lower_bound(d);
        if(j!=end() && j->first==d) return j;//d is the lh side of the closed interval
                                             //remember j->first >= d because it was lower but its the first
        if(j==begin()) return end();//d < all intervals
        --j;                        //back up
        return j;
    }
    const_iterator find(domain& d) const{
        iterator j = greatest_lower_bound(d);
        if (is_in(j,d)) return j;
        return end();
    }  //so much for find(d) 
    iterator find(domain& d){
        iterator j = greatest_lower_bound(d);
        if (is_in(j,d)) return j;
        return end();
    }  //so much for find(d) 

     struct overlap: public std::exception{
     };
     bool erase(const double lhep,const double rhep);
     //you have a lot of work regarding splitting intervals erasing when overlapped
     //but that can all be done with erase, and insert below. 
     //erase may need to split too
     std::pair<iterator,bool>
     split_and_or_erase_intervals(const double lhep,
                                  const double rhep, 
                                  const codomain& cd);
     //the insert method - note the addition of the overwrtite 
     std::pair<iterator,bool>
     insert(const double lhep,const double rhep,const codomain& cd,bool overwrite_ok){
          if( find(lhep)!=end() || find(rhep)!=end() ) {
              if(overwrite_ok){
                 return split_and_or_erase_intervals(const double lhep,
                                                     const double rhep, 
                                                     const codomain& cd);
              }
              throw overlap();
          }
          return insert(value_type(lhep,pair<double,codomain>(rhep,cd)));
     }
 };
于 2009-08-25T09:00:53.367 回答
0

If your intervals must be non-overlapping, you must add some extra code to verify this property at insertion-time. Specifically, the property you wish to assert is that your new interval lies entirely within a range that was previously empty. An easy way to do this is to allow two types of ranges: "occupied" and "empty". You should begin by creating a single "empty" entry which covers the entire usable range. Insertion of a new "occupied" range requires:

(1) lookup some value within the new range.
(2) ensure that the returned range is empty, and wholly encompasses your new range. (This was the required assertion, above)
(3) modify the returned empty range so its end lies at the start of your new range. (4) insert a new empty range that begins at the end of your new range, and ends at the old end of the returned range.
(5) insert your new range, confident that it is surrounded by empty-ranges.
(6) There may be additional corner-cases when inserting a new occupied range which has no empty space separating it from other occupied ranges.

于 2012-08-09T15:21:50.130 回答