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<!DOCTYPE html><html class="theme-next pisces use-motion" lang="zh-CN"><head><meta name="generator" content="Hexo 3.9.0"><meta charset="UTF-8"><meta http-equiv="X-UA-Compatible" content="IE=edge"><meta name="viewport" content="width=device-width,initial-scale=1,maximum-scale=2"><meta name="theme-color" content="#222"><link rel="apple-touch-icon" sizes="180x180" href="/images/icons/favicon-48.png?v=6.7.0"><link rel="icon" type="image/png" sizes="32x32" href="/images/icons/favicon-32.png?v=6.7.0"><link rel="icon" type="image/png" sizes="16x16" href="/images/icons/favicon-16.png?v=6.7.0"><link rel="mask-icon" href="/images/icons/logo.svg?v=6.7.0" color="#222"><link rel="alternate" href="/atom.xml" title="菠菜眾長" type="application/atom+xml"><meta http-equiv="Cache-Control" content="no-transform"><meta http-equiv="Cache-Control" content="no-siteapp"><link href="/lib/fancybox/source/jquery.fancybox.css" rel="stylesheet" type="text/css"><link 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id="posts" class="posts-expand"><div class="reading-progress-bar"></div><article class="post post-type-normal" itemscope itemtype="http://schema.org/Article"><div class="post-block"><link itemprop="mainEntityOfPage" href="https://lruihao.cn/posts/poj2387.html"><span hidden itemprop="author" itemscope itemtype="http://schema.org/Person"><meta itemprop="name" content="李瑞豪"><meta itemprop="description" content="从ACM到Web，分享程序、技巧、干货，记录心情、学习、成长！"><meta itemprop="image" content="/images/avatar.png"></span><span hidden itemprop="publisher" itemscope itemtype="http://schema.org/Organization"><meta itemprop="name" content="菠菜眾長"></span><header class="post-header"><h2 class="post-title" itemprop="name headline">Til the Cows Come Home-poj2387(dijkstra判断重边)<a href="https://github.com/Lruihao/lruihao.github.io/tree/hexo/source/_posts/poj2387.md" class="post-edit-link" title="编辑" rel="external nofollow noopener noreferrer" target="_blank"><i class="fa fa-pencil"></i></a></h2><div class="post-meta"><span class="post-time"><span class="post-meta-item-icon"><i class="fa fa-calendar-o"></i> </span><span class="post-meta-item-text">发表于</span> <time title="创建时间：2018-08-03 21:40:33" itemprop="dateCreated datePublished" datetime="2018-08-03T21:40:33+08:00">2018-08-03</time> </span><span class="post-category"><span class="post-meta-divider">|</span> <span class="post-meta-item-icon"><i class="fa fa-folder-o"></i> </span><span class="post-meta-item-text">分类于</span> <span itemprop="about" itemscope itemtype="http://schema.org/Thing"><a href="/categories/ACM/" itemprop="url" rel="index"><span itemprop="name">ACM</span></a></span> ， <span itemprop="about" itemscope itemtype="http://schema.org/Thing"><a href="/categories/ACM/最短路/" itemprop="url" rel="index"><span itemprop="name">最短路</span></a></span> </span><span class="post-comments-count"><span class="post-meta-divider">|</span> <span class="post-meta-item-icon"><i class="fa fa-comment-o"></i> </span><span class="post-meta-item-text">评论数： </span><a href="/posts/poj2387.html#comments" itemprop="discussionUrl"><span class="post-comments-count valine-comment-count" data-xid="/posts/poj2387.html" itemprop="commentCount"></span> </a></span><span id="/posts/poj2387.html" class="leancloud_visitors" data-flag-title="Til the Cows Come Home-poj2387(dijkstra判断重边)"><span class="post-meta-divider">|</span> <span title="阅读次数"><span class="post-meta-item-icon"><i class="fa fa-eye"></i> </span><span class="post-meta-item-text">阅读次数：</span> <span class="leancloud-visitors-count"></span></span></span><div class="post-symbolscount"><span class="post-meta-divider">|</span> <span class="post-meta-item-icon"><i class="fa fa-file-word-o"></i> </span><span class="post-meta-item-text">本文字数：</span> <span title="本文字数">412</span> <span class="post-meta-divider">|</span> <span class="post-meta-item-icon"><i class="fa fa-clock-o"></i> </span><span class="post-meta-item-text">阅读时长 ≈</span> <span title="阅读时长">1 分钟</span></div></div></header><div class="post-body" itemprop="articleBody"><p><a href="http://poj.org/problem?id=2387" rel="external nofollow noopener noreferrer" target="_blank">题目链接</a></p><p>题目大意：<br>说的是，一只奶牛位于N号节点，输入N个节点和T对双向的边，求出由N到1的最短的距离，其实就是问的单源最短路问题。</p><p>两个点可能有多条路，选择最短的。<br></p><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span><span class="meta-string"><stdio.h></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span><span class="meta-string"><string.h></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span><span class="meta-string"><algorithm></span></span></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> <span class="built_in">std</span>;</span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> INF=<span class="number">99999999</span>; <span class="comment">//设为无穷大</span></span><br><span class="line"><span class="keyword">int</span> maps[<span class="number">1005</span>][<span class="number">1005</span>],v[<span class="number">1005</span>],d[<span class="number">1005</span>]; <span class="comment">//v表示是否已经过遍历 d表示从源到点当前最短路</span></span><br><span class="line"><span class="keyword">int</span> n;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">Dijkstra</span><span class="params">(<span class="keyword">int</span> s,<span class="keyword">int</span> t)</span></span></span><br><span class="line"><span class="function"></span>{</span><br><span class="line"> <span class="keyword">int</span> i,j,k,mini;</span><br><span class="line"> <span class="keyword">for</span>(i=<span class="number">1</span>;i<=n;i++)</span><br><span class="line"> d[i]=INF; <span class="comment">//除源点设为0距离外 其他先设为无穷大</span></span><br><span class="line"> d[s]=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(i=<span class="number">1</span>;i<=n;i++) <span class="comment">//n点循环n次 ,找出n个k,找n个点</span></span><br><span class="line"> {</span><br><span class="line"> mini=INF;</span><br><span class="line"> k=<span class="number">-1</span>;</span><br><span class="line"> <span class="keyword">for</span>(j=<span class="number">1</span>;j<=n;j++) <span class="comment">//在所有未标记点中 选d值最小的点</span></span><br><span class="line"> <span class="keyword">if</span>(!v[j] && d[j]<mini)</span><br><span class="line"> mini=d[k=j];</span><br><span class="line"></span><br><span class="line"> v[k]=<span class="number">1</span>; <span class="comment">//标记节点</span></span><br><span class="line"> <span class="keyword">if</span>(k==t)</span><br><span class="line"> {</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,d[t]);</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(j=<span class="number">1</span>;j<=n;j++)</span><br><span class="line"> <span class="keyword">if</span>(!v[j] && (d[k]+maps[k][j])<d[j]) <span class="comment">//表示从k出发的点，对于所有边，更新相连点</span></span><br><span class="line"> d[j]=d[k]+maps[k][j];</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span></span><br><span class="line"><span class="function"></span>{</span><br><span class="line"> <span class="keyword">int</span> T,i,j,x,y,D;</span><br><span class="line"> <span class="keyword">while</span>(<span class="built_in">scanf</span>(<span class="string">"%d %d"</span>,&T,&n)!=EOF)</span><br><span class="line"> {</span><br><span class="line"> <span class="built_in">memset</span>(v,<span class="number">0</span>,<span class="keyword">sizeof</span>(v)); <span class="comment">//清除标记</span></span><br><span class="line"> <span class="keyword">for</span>(i=<span class="number">1</span>;i<=n;i++)</span><br><span class="line"> <span class="keyword">for</span>(j=<span class="number">1</span>;j<=n;j++)</span><br><span class="line"> maps[i][j]=INF;</span><br><span class="line"></span><br><span class="line"> <span class="keyword">for</span>(i=<span class="number">1</span>;i<=T;i++){</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d%d"</span>,&x,&y,&D);</span><br><span class="line"> <span class="keyword">if</span>(maps[x][y]>D) <span class="comment">//可能有多条路，只记录最短的</span></span><br><span class="line"> maps[x][y]=D,maps[y][x]=D;</span><br><span class="line"> }</span><br><span class="line"> Dijkstra(<span class="number">1</span>,n);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p></p></div><div 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}
// sort slices in content by search text's count and hits' count
slicesOfContent.sort(function (sliceLeft, sliceRight) {
if (sliceLeft.searchTextCount !== sliceRight.searchTextCount) {
return sliceRight.searchTextCount - sliceLeft.searchTextCount;
} else if (sliceLeft.hits.length !== sliceRight.hits.length) {
return sliceRight.hits.length - sliceLeft.hits.length;
} else {
return sliceLeft.start - sliceRight.start;
}
});
// select top N slices in content
var upperBound = parseInt('1');
if (upperBound >= 0) {
slicesOfContent = slicesOfContent.slice(0, upperBound);
}
// highlight title and content
function highlightKeyword(text, slice) {
var result = '';
var prevEnd = slice.start;
slice.hits.forEach(function (hit) {
result += text.substring(prevEnd, hit.position);
var end = hit.position + hit.length;
result += '<b class="search-keyword">' + text.substring(hit.position, end) + '</b>';
prevEnd = end;
});
result += text.substring(prevEnd, slice.end);
return result;
}
var resultItem = '';
if (slicesOfTitle.length != 0) {
resultItem += "<li><a href='" + articleUrl + "' class='search-result-title'>" + highlightKeyword(title, slicesOfTitle[0]) + "</a>";
} else {
resultItem += "<li><a href='" + articleUrl + "' class='search-result-title'>" + title + "</a>";
}
slicesOfContent.forEach(function (slice) {
resultItem += "<a href='" + articleUrl + "'>" +
"<p class=\"search-result\">" + highlightKeyword(content, slice) +
"...</p>" + "</a>";
});
resultItem += "</li>";
resultItems.push({
item: resultItem,
searchTextCount: searchTextCount,
hitCount: hitCount,
id: resultItems.length
});
}
})
};
if (keywords.length === 1 && keywords[0] === "") {
resultContent.innerHTML = '<div id="no-result"><i class="fa fa-search fa-5x" /></div>'
} else if (resultItems.length === 0) {
resultContent.innerHTML = '<div id="no-result"><i class="fa fa-frown-o fa-5x" /></div>'
} else {
resultItems.sort(function (resultLeft, resultRight) {
if (resultLeft.searchTextCount !== resultRight.searchTextCount) {
return resultRight.searchTextCount - resultLeft.searchTextCount;
} else if (resultLeft.hitCount !== resultRight.hitCount) {
return resultRight.hitCount - resultLeft.hitCount;
} else {
return resultRight.id - resultLeft.id;
}
});
var searchResultList = '<ul class=\"search-result-list\">';
resultItems.forEach(function (result) {
searchResultList += result.item;
})
searchResultList += "</ul>";
resultContent.innerHTML = searchResultList;
}
}
if ('auto' === 'auto') {
input.addEventListener('input', inputEventFunction);
} else {
$('.search-icon').click(inputEventFunction);
input.addEventListener('keypress', function (event) {
if (event.keyCode === 13) {
inputEventFunction();
}
});
}
// remove loading animation
$(".local-search-pop-overlay").remove();
$('body').css('overflow', '');
proceedsearch();
}
});
}
// handle and trigger popup window;
$('.popup-trigger').click(function(e) {
e.stopPropagation();
if (isfetched === false) {
searchFunc(path, 'local-search-input', 'local-search-result');
} else {
proceedsearch();
};
});
$('.popup-btn-close').click(onPopupClose);
$('.popup').click(function(e){
e.stopPropagation();
});
$(document).on('keyup', function (event) {
var shouldDismissSearchPopup = event.which === 27 &&
$('.search-popup').is(':visible');
if (shouldDismissSearchPopup) {
onPopupClose();
}
});</script><script>!function(){var t=document.createElement("script"),e=window.location.protocol.split(":")[0];t.src="https"===e?"https://zz.bdstatic.com/linksubmit/push.js":"http://push.zhanzhang.baidu.com/push.js";var s=document.getElementsByTagName("script")[0];s.parentNode.insertBefore(t,s)}()</script><script src="/lib/pangu/dist/pangu.min.js?v=3.3"></script><script type="text/javascript">pangu.spacingPage()</script><script src="/lib/bookmark/bookmark.min.js?v=1.0"></script><script type="text/javascript">bookmark.scrollToMark("manual","#更多")</script><script>!function(e){var r=Array.prototype.slice.call(document.querySelectorAll("img[data-original]"));function t(){for(var c=0;c<r.length;c++)t=r[c],void 0,0<=(n=t.getBoundingClientRect()).top&&0<=n.left&&n.top<=(e.innerHeight||document.documentElement.clientHeight)&&function(){var t,n,e,i,o=r[c];t=o,n=function(){r=r.filter(function(t){return o!==t})},e=new Image,i=t.getAttribute("data-original"),e.onload=function(){t.src=i,n&&n()},e.src=i}();var t,n}t(),e.addEventListener("scroll",function(){!function(t,n){clearTimeout(t.tId),t.tId=setTimeout(function(){t.call(n)},500)}(t,e)})}(this);</script></body></html>