1110题。南加州大学2012年秋季ACM:火星的恶作剧

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{“关系”:[{“relationshipType”:“http://schemas.mathworks.com/matlab/code/2013/relationships/document”,“targetMode”:“”,“relationshipId”:“rId1”,“目标”:“/ matlab / document.xml”}, {“relationshipType”:“http://schemas.mathworks.com/matlab/code/2013/relationships/output”,“targetMode”:“”,“relationshipId”:“rId2”,“目标”:“/ matlab / output.xml”}],“部分”:[{“partUri”:“/ matlab / document.xml”,“关系”:[],“contentType”:“应用程序/ vnd.mathworks.matlab.code.document + xml”,“内容”:“< ?xml version = = \ \ 1.0 \”编码“utf - 8 \”?> \ n < w:文档xmlns: w = \ " http://schemas.openxmlformats.org/wordprocessingml/2006/main \ " > < w:身体> < w: p > < w: pPr > < w: pStyle w: val = \ "文本\ " / > < / w: pPr > < w: r > < w: t >这一挑战是解决问题,火星恶作剧,< / w: t > < / w: r > < w: r > < w: t > < / w: t > < / w: r > < w w:超链接:docLocation = \ " http://contest.usc.edu/index.php/Fall12/Home \ " > < w: r > < w: t > ACM南加州大学2012年秋季比赛< / w: t > < / w: r > < / w:超链接> < w: r > < w: t >。< / w: t > < / w: r > < / w: p > < w: p > < w: pPr > < w: pStyle w: val = \ "文本\ " / > < / w: pPr > < w: r > < w: t >岩石之间移动探测器的照片。这证明火星上存在生命,但是生命有多快?< / w: t > < / w: r > < / w: p > < w: p > < w: pPr > < w: pStyle w: val = \ "文本\ " / > < / w: pPr > < w: r > < w: t >所有的岩石是由一个有界与相反的角落(0,0)和(1,1)。相机只能看到这个1 x1区域。一个火星岩石。火星旅行的最小距离1 x1边缘的岩石初始位置和最后一个岩石位置边缘隐藏在速度V移动时岩石马蒂的速度降低到V / 2。一个无动于衷的岩石允许马蒂副业。注意,相机软件不跟踪特定的岩石,因此一个无动于衷Pict1岩石可能项目1和这个相同的岩石可能第四项第二幅。多个岩石可能在一幅画中占据同一位置。</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Return the Minimum Martian Speed to transition from the initial Rock set to the Final rock set.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:rPr><w:b/></w:rPr><w:t>Input: [Rock_locations_Pict1_Pict2,Time between pictures ]</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:rPr><w:b/></w:rPr><w:t>Output: [Minimum Martian Speed]</w:t></w:r><w:r><w:t>; to 2 decimal places;</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:rPr><w:b/></w:rPr><w:t>Scoring: Cumulative Time ( msec )</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>The full</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://contest.usc.edu/index.php/Fall12/Home?action=download&upname=pranks.in.txt\"><w:r><w:t>USC data file</w:t></w:r></w:hyperlink></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:rPr><w:b/></w:rPr><w:t>Example:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:rPr><w:b/></w:rPr><w:t>Input: [0.3 0.6;0.4 0.5;0.5 0.5;0.95 0.2;0.6 0.5;0.9 0.4;0.5 0.5; 0.3 0.6], 3.0</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Essentially, 4 rocks in two pictures with a time between pictures of 3 seconds</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:rPr><w:b/></w:rPr><w:t>Output: 0.37</w:t></w:r><w:r><w:t> is quite obvious, not in the least</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>This answer comes by moving [I1 F4; I2 F3; I3 F1; I4 F2] with rows two and three being of equal maximum effective distance of 1.1. My \"effective distance\" for I2 F3 is .4 + (.1)*2 + .5, where .1 is hypot(I2,F3) and multiply by 2 for speed reduction. This 1.1/3 rounds to 0.37.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:hyperlink w:docLocation=\"http://contest.usc.edu/index.php/Fall12/Home?action=download&upname=pranks.zhencao.cpp.txt\"><w:r><w:t>The Winner's Pranks solution</w:t></w:r></w:hyperlink><w:r><w:t>. This solution is much tighter than my first method thus I am scoring by time. Hopefully someone can champion Matlab and produce a succinct and worthy solution. The winner solved this in 22 minutes and second place took a leisurely 65 minutes. Time Started when the page was opened.</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

这一挑战是解决问题,火星恶作剧,< a href = " http://contest.usc.edu/index.php/Fall12/Home " >的南加州大学ACM 2012年秋季比赛< / >。 岩石之间移动探测器的照片。这证明火星上存在生命,但是生命有多快? 所有的岩石是由一个有界与相反的角落(0,0)和(1,1)。相机只能看到这个1 x1区域。一个火星岩石。火星旅行的最小距离1 x1边缘的岩石初始位置和最后一个岩石位置边缘隐藏在速度V移动时岩石马蒂的速度降低到V / 2。一个无动于衷的岩石允许马蒂副业。注意,相机软件不跟踪特定的岩石,因此一个无动于衷Pict1岩石可能项目1和这个相同的岩石可能第四项第二幅。多个岩石可能在一幅画中占据同一位置。 返回最低火星速度过渡从最初的岩石将最终的岩石集。 输入:[Rock_locations_Pict1_Pict2,间隔时间图片] 输出:[最低火星速度]; to 2 decimal places;Scoring: Cumulative Time ( msec )The full <a href="http://contest.usc.edu/index.php/Fall12/Home?action=download&upname=pranks.in.txt">USC data file</a>Example:Input: [0.3 0.6;0.4 0.5;0.5 0.5;0.95 0.2;0.6 0.5;0.9 0.4;0.5 0.5; 0.3 0.6], 3.0Essentially, 4 rocks in two pictures with a time between pictures of 3 secondsOutput: 0.37 is quite obvious, not in the leastThis answer comes by moving [I1 F4; I2 F3; I3 F1; I4 F2] with rows two and three being of equal maximum effective distance of 1.1. My "effective distance" for I2 F3 is .4 + (.1)*2 + .5, where .1 is hypot(I2,F3) and multiply by 2 for speed reduction. This 1.1/3 rounds to 0.37.<a href="http://contest.usc.edu/index.php/Fall12/Home?action=download&upname=pranks.zhencao.cpp.txt">The Winner's Pranks solution</a>. This solution is much tighter than my first method thus I am scoring by time. Hopefully someone can champion Matlab and produce a succinct and worthy solution. The winner solved this in 22 minutes and second place took a leisurely 65 minutes. Time Started when the page was opened.

这个挑战是解决问题,火星恶作剧,< http://contest.usc.edu/index.php/Fall12/Home ACM南加州大学2012年秋季比赛>。岩石之间移动探测器的照片。这证明火星上存在生命,但是生命有多快?岩石都是有界的广场对面的角落(0,0)和(1,1)。相机只能看到这个1 x1区域。一个火星岩石。火星旅行的最小距离1 x1边缘的岩石初始位置和最后一个岩石位置边缘隐藏在速度V移动时岩石马蒂的速度降低到V / 2。一个无动于衷的岩石允许马蒂副业。注意,相机软件不跟踪特定的岩石,因此一个无动于衷Pict1岩石可能项目1和这个相同的岩石可能第四项第二幅。多个岩石可能在一幅画中占据同一位置。 Return the Minimum Martian Speed to transition from the initial Rock set to the Final rock set. *Input: [Rock_locations_Pict1_Pict2,Time between pictures ]* *Output: [Minimum Martian Speed]*; to 2 decimal places; *Scoring: Cumulative Time ( msec )* The full <http://contest.usc.edu/index.php/Fall12/Home?action=download&upname=pranks.in.txt USC data file> *Example:* *Input: [0.3 0.6;0.4 0.5;0.5 0.5;0.95 0.2;0.6 0.5;0.9 0.4;0.5 0.5; 0.3 0.6], 3.0* Essentially, 4 rocks in two pictures with a time between pictures of 3 seconds *Output: 0.37* is quite obvious, not in the least This answer comes by moving [I1 F4; I2 F3; I3 F1; I4 F2] with rows two and three being of equal maximum effective distance of 1.1. My "effective distance" for I2 F3 is .4 + (.1)*2 + .5, where .1 is hypot(I2,F3) and multiply by 2 for speed reduction. This 1.1/3 rounds to 0.37. <http://contest.usc.edu/index.php/Fall12/Home?action=download&upname=pranks.zhencao.cpp.txt The Winner's Pranks solution>. This solution is much tighter than my first method thus I am scoring by time. Hopefully someone can champion Matlab and produce a succinct and worthy solution. The winner solved this in 22 minutes and second place took a leisurely 65 minutes. Time Started when the page was opened.

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1110题。南加州大学2012年秋季ACM:火星的恶作剧

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