传送门 Find them, Catch them Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 42463 Accepted: 13065 Description The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang D…
题目链接: K - Find them, Catch them POJ - 1703 题目大意:警方决定捣毁两大犯罪团伙:龙帮和蛇帮,显然一个帮派至少有一人.该城有N个罪犯,编号从1至N(N<=100000.将有M(M<=100000)次操作.D a b 表示a.b是不同帮派A a b 询问a.b关系. 具体思路:带权并查集模板题.一般并查集都是相同的放在一个联通块里面.对于这个题,我们可以利用一下这个性质,只要是有联系的,都放进一个连通块里面,然后我们查询的时候,如果说他们的祖先一样的话,就…
Description The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present…
The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question…
题意: 有两个黑社会帮派,有n个人,他们肯定属于两个帮派中的一个,然后有两种操作 1 D a b 给出a b 两个人不属于同一个帮派 2 A a b 问a b 两个人关系 输出 同一个帮派,不是同一个帮派,或者不确定 思路: 比较简单的带权并查集,容易想,方法固定,可以开距离根节点的距离这个权,或者是异或也行(应该是行,没试),如果是距离根节点的话 就是每次D a b 都把a b 看成是距离1,然后接在一起,A a b 时候看是不是同一个祖先,是的话看看距离权值是否奇…
True Liars Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 2713 Accepted: 868 Description After having drifted about in a small boat for a couple of days, Akira Crusoe Maeda was finally cast ashore on a foggy island. Though he was exha…
Navigation Nightmare Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 5939 Accepted: 2102 Case Time Limit: 1000MS Description Farmer John's pastoral neighborhood has N farms (2 <= N <= 40,000), usually numbered/labeled 1..N. A series o…