4

我需要使用堆,所以我搜索了 STL,但它似乎不起作用,我写了一些代码来解释我的意思:

#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>

struct data
{
    int indice;
    int tamanho;
};


bool comparator2(const data* a, const data* b)
{
    return (a->tamanho < b->tamanho);
}

int main()
{

        std::vector<data*> mesas;

        data x1, x2, x3, x4, x5;

        x1.indice = 1;
        x1.tamanho = 3;

        x2.indice = 2;
        x2.tamanho = 5;

        x3.indice = 3;
        x3.tamanho = 2;

        x4.indice = 4;
        x4.tamanho = 6;

        x5.indice = 5;
        x5.tamanho = 4;

        mesas.push_back(&x1);

        mesas.push_back(&x2);

        mesas.push_back(&x3);

        mesas.push_back(&x4);

        mesas.push_back(&x5);


        make_heap(mesas.begin(), mesas.end(), comparator2);

        for(int i = 0 ; i < 5 ; i++)
        {
            data* mesa = mesas.front();
            pop_heap(mesas.begin(),mesas.end());
            mesas.pop_back();

            printf("%d, %d\n", mesa->indice, mesa->tamanho);
        }

    return 0;
};

这就是我得到的:

4, 6
2, 5
1, 3
3, 2
5, 4

所以它不能作为堆工作,因为向量上的最大元素没有被正确返回。

还是我做错了什么?

4

4 回答 4

14

您需要传递comparator2给,std::pop_heap因为这就是您创建堆的方式。否则,它将对指针使用默认的小于运算符,它只是比较指针值。

于 2010-05-11T21:28:44.307 回答
1

MSN 的回答是正确的。但是,几个样式指南中的任何一个都可以防止此错误:

  • 声明比较器处理引用,而不是对象operator<。使用vector对象,而不是指针。

    bool comparator2(const data& a, const data& b)
    {
        return (a.tamanho < b.tamanho);
    }
    

    您可能真的需要指针向量,在这种情况下这不适用。

  • 使用std::priority_queue(from <queue>),它联系在一起pop_heappop_back为你,记住你的比较器。这需要一个函子比较器:

    struct comparator2 { bool operator()(const data& a, const data& b)
    {
        return (a.tamanho < b.tamanho);
    } };
     
    std::priority_queue<data, vector<data>, comparator2> mesas;
     // or std::priority_queue<data, vector<data>, comparator2>
    mesas.push(x1);
    
  • 最优雅的方法是将此作为默认比较data

    struct data
    {
        int indice;
        int tamanho;
         
        friend bool operator<(const data& a, const data& b)
        {
            return (a.tamanho < b.tamanho);
        }
    };
    std::priority_queue<data> mesas;
    mesas.push(x1);
    

priority_queue也可以采用预填充的未排序容器,它将复制。

于 2010-05-11T22:22:47.257 回答
0

我遇到了类似的问题,并且能够通过以下方式解决它:

struct comparator2 { bool operator()(data const * const a, data const * const b)
{
    return (a->tamanho < b->tamanho);
 } };

std::priority_queue<data*, std::vector<data*>, comparator2> mesas;
于 2013-11-25T05:10:49.517 回答
0

std::set 怎么样

#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <set>

struct data
{
    // Always put constructors on.
    // When the constructor is finished the object is ready to be used.
    data(int i,int t)
        :indice(i)
        ,tamanho(t)
    {}

    int indice;
    int tamanho;

    // Add the comparator to the class.
    // Then you know where to look for it.
    bool operator<(data const& b) const
    {
        return (tamanho < b.tamanho);
    }
};



int main()
{

        std::set<data> mesas;

        // Dont declare all your variables on the same line.
        // One per line otherwise it is hard to read.
        data x1(1,3);
        data x2(2,5);
        data x3(3,2);
        data x4(4,6);
        data x5(5,4);

        mesas.insert(x1);
        mesas.insert(x2);
        mesas.insert(x3);
        mesas.insert(x4);
        mesas.insert(x5);
        // You don't actually need the variables.
        // You could have done it in place.
        mesas.insert(data(6,100));

        // Use iterator to loop over containers.
        for(std::set<data>::iterator loop = mesas.begin(); loop != mesas.end(); ++loop)
        {
            printf("%d, %d\n", loop->indice, loop->tamanho);
        }

    return 0;
};
于 2010-05-11T22:03:10.320 回答