Rivers, floods, and the rhythm of spring

Robby Robin's Journey

Earlier today, I joined the throngs of fellow citizens of our fair city out walking and biking in the spring sunshine. A leisurely stroll or ride along the river and around downtown is always enjoyable on a sunny Sunday, but today everyone had something special in mind. Everyone was out viewing the high waters of this year’s spring freshet. And since this year’s flood has caused the closure of several downtown streets to cars due to high water, the neighbourhoods near the river are wonderfully quiet, with only the sounds of people enjoying each other’s company and the occasional sound of a sump pump motor sucking water from someone’s basement.

Downtown Fredericton, Library on right

Floods aren’t fun. They damage property, dislocate people and animals, inconvenience countless others, cost lots of money, and in the worst cases can result in staggering human tragedy. They are also regular occurrences, so why have…

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【转载】union-find 算法总结


基础版就是union-find,进阶的几个基本都是用了树这种数据结构,包括quick-union、加权quick-union,路径压缩的加权quick-union。

1.
https://senlinzhan.github.io/2015/01/14/unionfind%E7%AE%97%E6%B3%95/

2.

https://blog.csdn.net/dm_vincent/article/details/7655764

3.
https://blog.csdn.net/yaokai_assultmaster/article/details/78888846

4.

https://www.cnblogs.com/SeaSky0606/p/4752941.html

女权主义与自由主义

最近看了些文章,隐隐有些感觉现在女权主义中的常见观点基本都是自由主义下面的,过去的文化女权主义等等式微了(更不行)。感觉还是因为自由主义更能说得通,而且覆盖面大(能解释的事物多)。美国70年代的女权运动是否也是趁着60年代黑人人权运动的势呢?美国的女权主义运动除了第一波之外(suffocate)时间上也较为重合。这也是为何女权主义在光谱中更偏左吧……

抛开短期的政治法律来说,一个主义(-ism)能否长期发展/比较火看的应该还是相关人士的智识。

最近看的文章中的其中一篇:

田雷|契约抑或身份?——劳动法在20世纪美国的兴起与衰落

https://mp.weixin.qq.com/s/8nMRWWoo8lQa9oMHH2Ctzg

【转载】binary tree inorder traversal

inorder 中序遍历

/**
*
* @param root 树根节点
* 利用栈模拟递归过程实现循环中序遍历二叉树
* 思想和上面的preOrderStack_2相同,只是访问的时间是在左子树都处理完直到null的时候出栈并访问。
*/
public static void inOrderStack(Node root){
if(root==null)return;
Stack s=new Stack();
while(root!=null||!s.isEmpty()){
while(root!=null){
s.push(root);//先访问再入栈
root=root.left;
}
root=s.pop();
System.out.println(root.value);
root=root.right;//如果是null,出栈并处理右子树
}
}