
并查集
gtuif
大多数人永远都不会了解真实的你,但是他们会看到你所做的一切。。。
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并查集详解(转)
来看一个实例,杭电1232畅通工程首先在地图上给你若干个城镇,这些城镇都可以看作点,然后告诉你哪些对城镇之间是有道路直接相连的。最后要解决的是整幅图的连通性问题。比如随意给你两个点,让你判断它们是否连通,或者问你整幅图一共有几个连通分支,也就是被分成了几个互相独立的块。像畅通工程这题,问还需要修几条路,实质就是求有几个连通分支。如果是1个连通分支,说明整幅图上的点都连起来了,不用再修路转载 2017-02-06 14:42:39 · 207 阅读 · 0 评论 -
The Suspects
Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to s原创 2017-02-09 22:51:37 · 278 阅读 · 0 评论 -
POJ-2524 Ubiquitous Religions
Ubiquitous ReligionsTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 33179 Accepted: 16090DescriptionThere are so many different religions in the world t原创 2017-02-09 23:09:15 · 220 阅读 · 0 评论 -
POJ-1182 食物链(经典并查集)
食物链Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 67377 Accepted: 19903Description动物王国中有三类动物A,B,C,这三类动物的食物链构成了有趣的环形。A吃B, B吃C,C吃A。 现有N个动物,以1-N编号。每个动物都转载 2017-02-09 22:26:48 · 320 阅读 · 0 评论 -
POJ-2236 Wireless Network
Wireless NetworkTime Limit: 10000MS Memory Limit: 65536KTotal Submissions: 25451 Accepted: 10574DescriptionAn earthquake takes place in Southeast Asia. The ACM (A原创 2017-02-11 15:09:51 · 197 阅读 · 0 评论 -
并查集知识总结(师傅的)
1.非路径压缩: 递归版:__int64 findroot(__int64 x){ if(x!=fa[x]) return findroot(fa[x]); return fa[x];}1234512345非递归版:__int64 findroot(__int64 x)//查找x的根节点{ __int64 r=x; while(r!=fa[转载 2017-02-13 11:22:28 · 154 阅读 · 0 评论 -
HDUOJ-3047 Zjinu Stadium
Zjnu StadiumTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3242 Accepted Submission(s): 1250Problem DescriptionIn 12th Zheji原创 2017-02-16 09:31:03 · 219 阅读 · 0 评论 -
How Many Answers Are Wrong
TT and FF are ... friends. Uh... very very good friends -________-b FF is a bad boy, he is always wooing TT to play the following game with him. This is a very humdrum game. To begin with, TT shou原创 2017-02-16 10:08:31 · 215 阅读 · 0 评论