Ramer-Douglas-Peucker路径简化算法

时间:2011-06-26 10:37:29

标签: c++ algorithm debugging

我在阅读完文章后实现了路径简化算法:

http://losingfight.com/blog/2011/05/30/how-to-implement-a-vector-brush/

对我来说,为我的游戏生成优化的关卡几何体非常有用。但是,我现在正在使用它来清理一个*寻路路径,并且它有一个奇怪的边缘情况,但却失败了。

以下是它工作的截图 - 优化从红色圆圈到蓝色圆圈的路径。淡绿色线是a *输出,浅白色线是优化路径。

enter image description here

这是一个失败的屏幕截图:

enter image description here

这是我的代码。我将文章中的ObjC代码改编为c ++

注意:vec2fvecstd::vector< vec2<float> >,'real'只是一个typedef'd float。

       void rdpSimplify( const vec2fvec &in, vec2fvec &out, real threshold )
{
    if ( in.size() <= 2 )
    {
        out = in;
        return;
    }

    //
    //  Find the vertex farthest from the line defined by the start and and of the path
    //

    real maxDist = 0;
    size_t maxDistIndex = 0;      
    LineSegment line( in.front(), in.back() );

    for ( vec2fvec::const_iterator it(in.begin()),end(in.end()); it != end; ++it )
    {
        real dist = line.distance( *it );
        if ( dist > maxDist )
        {
            maxDist = dist;
            maxDistIndex = it - in.begin();
        }
    }

    //
    //  If the farhtest vertex is greater than our threshold, we need to
    //  partition and optimize left and right separately
    //

    if ( maxDist > threshold )
    {
        //
        //  Partition 'in' into left and right subvectors, and optimize them
        //

        vec2fvec left( maxDistIndex+1 ),
                 right( in.size() - maxDistIndex ),
                 leftSimplified,
                 rightSimplified;

        std::copy( in.begin(), in.begin() + maxDistIndex + 1, left.begin() );
        std::copy( in.begin() + maxDistIndex, in.end(), right.begin() );

        rdpSimplify(left, leftSimplified, threshold );
        rdpSimplify(right, rightSimplified, threshold );

        //
        //  Stitch optimized left and right into 'out'
        //

        out.resize( leftSimplified.size() + rightSimplified.size() - 1 );
        std::copy( leftSimplified.begin(), leftSimplified.end(), out.begin());
        std::copy( rightSimplified.begin() + 1, rightSimplified.end(), out.begin() + leftSimplified.size() );
    }
    else
    {
        out.push_back( line.a );
        out.push_back( line.b );
    }
}

我真的不知道出了什么问题。我的蜘蛛侠意识说它是在std :: copy调用中......我必须在某些情况下复制垃圾。

编辑: 我已经重写了算法,不再使用迭代器和std :: copy等。它仍然以完全相同的方式失败。

       void rdpSimplify( const vec2fvec &in, vec2fvec &out, real threshold )
{
    if ( in.size() <= 2 )
    {
        out = in;
        return;
    }

    //
    //  Find the vertex farthest from the line defined by the start and and of the path
    //

    real maxDist = 0;
    size_t maxDistIndex = 0;      
    LineSegment line( in.front(), in.back() );

    for ( size_t i = 0, N = in.size(); i < N; i++ )
    {
        real dist = line.distance( in[i] );
        if ( dist > maxDist )
        {
            maxDist = dist;
            maxDistIndex = i;
        }
    }


    //
    //  If the farthest vertex is greater than our threshold, we need to
    //  partition and optimize left and right separately
    //

    if ( maxDist > threshold )
    {
        //
        //  Partition 'in' into left and right subvectors, and optimize them
        //


        vec2fvec left, right, leftSimplified, rightSimplified;
        for ( size_t i = 0; i < maxDistIndex + 1; i++ ) left.push_back( in[i] );
        for ( size_t i = maxDistIndex; i < in.size(); i++ ) right.push_back( in[i] );


        rdpSimplify(left, leftSimplified, threshold );
        rdpSimplify(right, rightSimplified, threshold );

        //
        //  Stitch optimized left and right into 'out'
        //

        out.clear();
        for ( size_t i = 0, N = leftSimplified.size(); i < N; i++ ) out.push_back(leftSimplified[i]);
        for ( size_t i = 1, N = rightSimplified.size(); i < N; i++ ) out.push_back( rightSimplified[i] );
    }
    else
    {
        out.push_back( line.a );
        out.push_back( line.b );
    }
}

1 个答案:

答案 0 :(得分:4)

我在代码中找不到任何错误。

有些事情要尝试:

  • 添加一些调试打印语句以检查失败案例中maxDist的内容。它应该非常低,但如果它出现高,那么你就知道你的线段距离代码存在问题。
  • 检查您看到的路径是否与算法返回的路径匹配。如果没有,那么你的路径渲染可能有问题吗?当路径只有两个点时,也许是一个错误?
  • 通过在算法开头打印出所有坐标,检查输入路径是否符合预期。

如果你只是调查一下,找到问题的原因不应该花太长时间。几分钟后,盯着代码是一种非常糟糕的调试方式。