sdjzujxc的专栏

Problem Description为了训练小希的方向感，Gardon建立了一座大城堡，里面有N个房间(N和M条通道(M，每个通道都是单向的，就是说若称某通道连通了A房间和B房间，只说明可以通过这个通道由A房间到达B房间，但并不说明通过它可以由B房间到达A房间。Gardon需要请你写个程序确认一下是否任意两个房间都是相互连通的，即：对于任意的i和j，至少存在一条路径可以从房间i到房间j，也存

The King’s ProblemTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1091    Accepted Submission(s): 404Problem Description In the

SPFTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4088 Accepted: 1889DescriptionConsider the two networks shown below. Assuming that data moves around th

Redundant PathsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6723 Accepted: 2924DescriptionIn order to get from one of the F (1 <= F <= 5,000) grazing f

NetworkTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7272 Accepted: 3415DescriptionA Telephone Line Company (TLC) is establishing a new telephone cable

Road ConstructionTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6503 Accepted: 3271DescriptionIt's almost summer time, and that means that it's almos

The Bottom of a GraphTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 7103 Accepted: 2907DescriptionWe will use the following (standard) definitions from g

求强连通分量的tarjan算法强连通分量：是有向图中的概念，在一个图的子图中，任意两个点相互可达，也就是存在互通的路径，那么这个子图就是强连通分量。（如果一个有向图的任意两个点相互可达，那么这个图就称为强连通图）。如果u是某个强连通分量的根，那么：（1）u不存在路径可以返回到它的祖先。（2）u的子树也不存在路径可以返回到u的祖先。· 例如：· 强连通分量。在一个非强连通图中

Popular CowsTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 18198 Accepted: 7330DescriptionEvery cow's dream is to become the most popular cow in the herd

Going from u to v or from v to u?Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 11871 Accepted: 3018DescriptionIn order to make their sons brave, Jiajia

Countries in WarTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1725 Accepted: 550DescriptionIn the year 2050, after different attempts of the UN to mai

Summer HolidayTime Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 976    Accepted Submission(s): 415Problem DescriptionTo see a World i

Proving EquivalencesTime Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1426    Accepted Submission(s): 535Problem DescriptionConsider t

题意：网络中的一学校可以将软件发送给其他一些学校，能够发送给谁取决于他们各自维护的一个清单。将学校看成一个节点，给出每个学校的维护清单，问至少需要复制几次软件，使毎个学校都能够得到该软件，在清单中至少添加几项，可使软件至少复制一次，所有学校都可以得到。思路：1、用Tarjan算法求出强连通分分量。2、缩点重新构图。3、分别求节点的出度和入度。第一个问题就是入度为零的个数，第二问题就是出

RevolC FaeLoNHopefully, you can remember Koorosh from one of the previous problems. Fortunately, Koorosh has solved Basm’s problem without your help! Now, he is a high-ranked advisor of Basm. As u