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<meta name="description" content="今天总结一下广度优先搜索(BFS). BFS是树/图的遍历的常用算法之一, 对于没有边权重的图来说可以计算最短路径. 由于树的BFS只是图的BFS的一种特殊情况, 而且比较简单不需要visited标记, 这里只写一下图的BFS好了. 先定义一个Graph类, 这里在每一个节点保存邻居信息: public class GraphNode{ int val; List<GraphNode> neighbors; } BFS for trees/graphs 图的遍历需要注意不走重复节点, 所以需要一个HashSet(名字叫visited)来保存哪些节点已经访问过了. 需要注意的是, 在把一个节点放进队列queue的时刻就要把它放进visited, 而不是在队列里取出来的时刻再放. public void BFS(GraphNode start){ LinkedList<GraphNode> q = new LinkedList<GraphNode>(); HasheSet<GraphNode> visited = new HasheSet<GraphNode>(); q.push(start); visited ..." />
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<meta property="og:description" content="今天总结一下广度优先搜索(BFS). BFS是树/图的遍历的常用算法之一, 对于没有边权重的图来说可以计算最短路径. 由于树的BFS只是图的BFS的一种特殊情况, 而且比较简单不需要visited标记, 这里只写一下图的BFS好了. 先定义一个Graph类, 这里在每一个节点保存邻居信息: public class GraphNode{ int val; List<GraphNode> neighbors; } BFS for trees/graphs 图的遍历需要注意不走重复节点, 所以需要一个HashSet(名字叫visited)来保存哪些节点已经访问过了. 需要注意的是, 在把一个节点放进队列queue的时刻就要把它放进visited, 而不是在队列里取出来的时刻再放. public void BFS(GraphNode start){ LinkedList<GraphNode> q = new LinkedList<GraphNode>(); HasheSet<GraphNode> visited = new HasheSet<GraphNode>(); q.push(start); visited ..."/>
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广度优先搜索(BFS)小结
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目录
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<div id="toc"><ul><li><a class="toc-href" href="#bfs-for-treesgraphs" title="BFS for trees/graphs">BFS for trees/graphs</a></li><li><a class="toc-href" href="#bfs-with-distance" title="BFS with distance">BFS with distance</a></li><li><a class="toc-href" href="#properties" title="properties">properties</a></li><li><a class="toc-href" href="#bfs-by-layer" title='BFS "by layer"'>BFS "by layer"</a></li><li><a class="toc-href" href="#complexity" title="complexity">complexity</a></li></ul></div>
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</div>
<p>今天总结一下广度优先搜索(BFS). BFS是树/图的遍历的常用算法之一, 对于没有边权重的图来说可以计算最短路径. <br/>
由于树的BFS只是图的BFS的一种特殊情况, 而且比较简单不需要visited标记, 这里只写一下图的BFS好了. <br/>
先定义一个Graph类, 这里在每一个节点保存邻居信息: </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">public class GraphNode{ </span></span>
<span class="code-line"><span class="err"> int val; </span></span>
<span class="code-line"><span class="err"> List<GraphNode> neighbors; </span></span>
<span class="code-line"><span class="err">}</span></span>
</pre></div>
<h2 id="bfs-for-treesgraphs">BFS for trees/graphs</h2>
<p>图的遍历需要注意不走重复节点, 所以需要一个HashSet(名字叫visited)来保存哪些节点已经访问过了. 需要注意的是, <em>在把一个节点放进队列queue的时刻就要把它放进visited</em>, 而不是在队列里取出来的时刻再放. </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">public void BFS(GraphNode start){ </span></span>
<span class="code-line"><span class="err"> LinkedList<GraphNode> q = new LinkedList<GraphNode>(); </span></span>
<span class="code-line"><span class="err"> HasheSet<GraphNode> visited = new HasheSet<GraphNode>(); </span></span>
<span class="code-line"><span class="err"> q.push(start); </span></span>
<span class="code-line"><span class="err"> visited.add(start); </span></span>
<span class="code-line"><span class="err"> while(!q.empty()){ </span></span>
<span class="code-line"><span class="err"> GraphNode cur = q.poll(); </span></span>
<span class="code-line"><span class="err"> System.out.println(cur.val); </span></span>
<span class="code-line"><span class="err"> for(GraphNode next: cur.children){ </span></span>
<span class="code-line"><span class="err"> if(!visited.contains(next)){ </span></span>
<span class="code-line"><span class="err"> q.push(next); </span></span>
<span class="code-line"><span class="err"> visited.add(next); // mark node as visited when adding to queue! </span></span>
<span class="code-line"><span class="err"> } </span></span>
<span class="code-line"><span class="err"> } </span></span>
<span class="code-line"><span class="err"> }//while </span></span>
<span class="code-line"><span class="err">}</span></span>
</pre></div>
<h2 id="bfs-with-distance">BFS with distance</h2>
<p>在BFS的同时我们可以记录从start节点到当前node的距离, 方法是把一个距离信息同时入队(封装一个<code>Pair<GraphNode, Integer></code>), 或者使用一个与queue<em>平行</em>的队列保存距离信息. <br/>
在上面的代码中, 加入: </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">//... </span></span>
<span class="code-line"><span class="err">LinkedList<Integer> distq = new LinkedList<Integer>(); </span></span>
<span class="code-line"><span class="err">distq.push(0);// distance from start to start </span></span>
<span class="code-line"><span class="err">//... </span></span>
<span class="code-line"><span class="err">// in the while(!q.empty()) loop: </span></span>
<span class="code-line"><span class="err"> int d = distq.poll();//get distance from start to current node </span></span>
<span class="code-line"><span class="err"> for(GraphNode next: node.children){ </span></span>
<span class="code-line"><span class="err"> distq.push(d+1);// distance from start to next node </span></span>
<span class="code-line"><span class="err"> //...</span></span>
</pre></div>
<p>对于Tree的情况来说, 这里的dist其实就是当前节点的深度depth. </p>
<h2 id="properties">properties</h2>
<p><strong>性质1:</strong> <br/>
每个节点node的distance都是node距离起始点start的最短距离. </p>
<p><strong>性质2:</strong> <br/>
距离start近的节点(depth浅的节点)一定比距离start远的节点早被访问到. </p>
<p>这是对一个树BFS的时候节点的访问顺序: <br/>
<img alt="" class="img-responsive" src="../images/bfs-summary/pasted_image.png"/> </p>
<h2 id="bfs-by-layer">BFS "by layer"</h2>
<p>参考上面的性质, 可以一次处理"一层"的节点, "一层"的意思是指所有节点距离start的距离相同. 代码在while循环里不是一次poll一个节点, 而是一次把queue的内容处理完, 然后换新的queue进入下一次while循环. 代码重新写一下: </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">public void BFS(GraphNode start){ </span></span>
<span class="code-line"><span class="err"> ArrayList<GraphNode> q = new ArrayList<Tree>(); </span></span>
<span class="code-line"><span class="err"> HasheSet<GraphNode> visited = new HasheSet<GraphNode>(); </span></span>
<span class="code-line"><span class="err"> q.push(start); </span></span>
<span class="code-line"><span class="err"> visited.add(start); </span></span>
<span class="code-line"><span class="err"> while(!q.empty()){ </span></span>
<span class="code-line"><span class="err"> ArrayList<GraphNode> newq = new ArrayList<Tree>();// create a new queue </span></span>
<span class="code-line"><span class="err"> for(GraphNode cur: q){// deal with all nodes in the queue </span></span>
<span class="code-line"><span class="err"> System.out.print(cur.val+", ");// all nodes in q are of the same distance/depth </span></span>
<span class="code-line"><span class="err"> for(GraphNode next: cur.children) </span></span>
<span class="code-line"><span class="err"> if(!visited.contains(next)) </span></span>
<span class="code-line"><span class="err"> { newq.push(next);visited.add(next); } </span></span>
<span class="code-line"><span class="err"> } </span></span>
<span class="code-line"><span class="err"> System.out.println(); </span></span>
<span class="code-line"><span class="err"> q = newq;//replace q with newq </span></span>
<span class="code-line"><span class="err"> }//while </span></span>
<span class="code-line"><span class="err">}</span></span>
</pre></div>
<p>以上程序每次打印一行, 第i行包括了距start距离为i的所有节点. <br/>
由于这样的话每次不必在队首poll出元素(而是依次处理所有queue的元素), 所以可以改用ArrayList. 此时while循环里的不变量是: 所有q里面的节点距离start的距离都相同. </p>
<h2 id="complexity">complexity</h2>
<p>假设一个图有N个节点和M条边, BFS会走遍所有节点, 时间是O(N), 然后由于每个节点会检查所有的出边, 最终所有的边都会被检查过, 时间是O(M), 所以BFS的时间复杂度是<strong>O(N+M)</strong>. </p>
<p>队列里面最多可能存放所有节点, 空间复杂度为<strong>O(N)</strong>. </p>
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$(document).ready(function() {
$.material.init();
$("div.navbar").autoHidingNavbar();
$(".img-responsive").css("cursor","pointer").on('click',function(){
var sr=$(this).attr('src');
$('#mimg').attr('src',sr);
$('#myModal').modal('show');
});
$('#affix-toc').affix({
offset: {
top: function(){
if($('#affix-toc').hasClass("affix"))
return $('#sidebar').height();
return $('#sidebar').height() - $('#affix-toc').height();
},
bottom: function (){
return $("#fcfooter").offset().top -
$("#article-content").offset().top -
$("#article-content").height() + 20;
}
}
});
$('#affix-toc').width($('#sidebar').width());
});
$(window).resize(function () {
$('#affix-toc').width($('#sidebar').width());
});
</script>
</body>
</html>