请给我一些关于如何评估覆盖所有给定区域的最大区域的建议？

发布于 2024-12-07 02:07:22 字数 544 浏览 0 评论 0原文

假设我有几十个地理区域，可以通过使用以下 c/c++ 结构来定义：

typedef struct tagGEOGRAPHIC_REGION
{    
    float fNorthMost;
    float fSouthMost;
    float fWestMost;
    float fEastMost;
} GEOGRAPHIC_REGION, *PGEOGRAPHIC_REGION;

现在我想要获得最大区域，它将覆盖所有给定区域。函数模板可能如下所示：

const GEOGRAPHIC_REGION& GetMaxRegion(const std:vector<GEOGRAPHIC_REGION>& vRegions)
{  
    ......
}

我可以将 GEOGRAPHIC_REGION 结构体的四个组成部分放入 4 个不同的浮点向量中，然后计算它们各自的最大值。最后，可以将四个最大值组合起来形成最大值区域。我认为这一定是一个简单的方法。您能给我一些建议吗？非常感谢！

原文

Suppose I have dozens of geographic regions, which can be defined through the use of the following c/c++ structure:

typedef struct tagGEOGRAPHIC_REGION
{    
    float fNorthMost;
    float fSouthMost;
    float fWestMost;
    float fEastMost;
} GEOGRAPHIC_REGION, *PGEOGRAPHIC_REGION;

And I now want to get the maximum region, which will cover all given regions. The function template may look like the following:

const GEOGRAPHIC_REGION& GetMaxRegion(const std:vector<GEOGRAPHIC_REGION>& vRegions)
{  
    ......
}

I can put the four components of GEOGRAPHIC_REGION struct into 4 different float vectors, and then evaluate their respective maximum values. Finally, the four maxium values can be combined to form the maximum region. I think it must be a simple way to do that. Would you please give me some advice? Thank you very much!

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自由如风 2024-12-14 02:07:22

难道你不能只迭代向量，并仅用一个循环获得最大北、最小南等吗？

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回忆凄美了谁 2024-12-14 02:07:22

我建议以这样的方式作为起点：

实时观看：http://ideone.com/LQk8U
~~请注意，我可能猜错了你的轴方向（在这种情况下，你可能需要交换一些最小/最大）~~ 编辑根据评论修复“北是正，东为正”
对于向量，只需执行 std::acummulate(v.begin(), v.end(), v [0]....)

如果您的场景变得更复杂，请参阅 Boost 几何库

。

#include <iostream>
#include <numeric>

struct GEOGRAPHIC_REGION
{    
    float fNorthMost;
    float fSouthMost;
    float fWestMost;
    float fEastMost;
};

GEOGRAPHIC_REGION combine(const GEOGRAPHIC_REGION& accum, const GEOGRAPHIC_REGION& tocombine)
{
    GEOGRAPHIC_REGION combined = { 
        std::max(accum.fNorthMost, tocombine.fNorthMost),
        std::min(accum.fSouthMost, tocombine.fSouthMost),
        std::min(accum.fWestMost,  tocombine.fWestMost),
        std::max(accum.fEastMost,  tocombine.fEastMost)
    };
    return combined;
}

int main()
{
    const GEOGRAPHIC_REGION regions[] = 
    {
        { 2,-1,-1,1 },
        { 1,-2,-1,1 },
        { 1,-1,-2,1 },
        { 1,-1,-1,2 },
    };

    GEOGRAPHIC_REGION super = std::accumulate(regions, regions+4, regions[0], combine);

    std::cout << "{ " << super.fNorthMost << ", " 
                      << super.fSouthMost << ", "
                      << super.fWestMost << ", "
                      << super.fEastMost << " }" << std::endl;
}

I suggest something like this as a starting point:

see it live: http://ideone.com/LQk8U
~~note that I might have your axis direction guessed wrong (you might need to swap some min/max in that case)~~ Edit fixed according to comment "North is positive, and East is positive"
for a vector, just do std::acummulate(v.begin(), v.end(), v[0]....)

In case your scenarios get more complicated, see the Boost Geometry Library

#include <iostream>
#include <numeric>

struct GEOGRAPHIC_REGION
{    
    float fNorthMost;
    float fSouthMost;
    float fWestMost;
    float fEastMost;
};

GEOGRAPHIC_REGION combine(const GEOGRAPHIC_REGION& accum, const GEOGRAPHIC_REGION& tocombine)
{
    GEOGRAPHIC_REGION combined = { 
        std::max(accum.fNorthMost, tocombine.fNorthMost),
        std::min(accum.fSouthMost, tocombine.fSouthMost),
        std::min(accum.fWestMost,  tocombine.fWestMost),
        std::max(accum.fEastMost,  tocombine.fEastMost)
    };
    return combined;
}

int main()
{
    const GEOGRAPHIC_REGION regions[] = 
    {
        { 2,-1,-1,1 },
        { 1,-2,-1,1 },
        { 1,-1,-2,1 },
        { 1,-1,-1,2 },
    };

    GEOGRAPHIC_REGION super = std::accumulate(regions, regions+4, regions[0], combine);

    std::cout << "{ " << super.fNorthMost << ", " 
                      << super.fSouthMost << ", "
                      << super.fWestMost << ", "
                      << super.fEastMost << " }" << std::endl;
}

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