URAL 1221. Malevich Strikes Back!
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1221 1221. Malevich Strikes Back! Time limit: 1.0 second Memory limit: 64 MB After the greatest success of Malevich’s “Black Square&#…
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URAL 1156. Two Rounds
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1156 1156. Two Rounds Time limit: 2.0 second Memory limit: 64 MB There are two rounds in the Urals Championship. The competitors have to solve N …
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URAL 1495. One-two, One-two 2
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1495 1495. One-two, One-two 2 Time limit: 2.0 second Memory limit: 64 MB A year ago the famous gangster Vito Maretti woke up in the morning and r…
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URAL 1009. K-based Numbers
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1009 1009. K-based Numbers Time limit: 1.0 second Memory limit: 64 MB Let’s consider K-based numbers, containing exactly N digits. We define a nu…
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POJ 1631. Bridging signals
1. 题目 http://poj.org/problem?id=1631 Bridging signals Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12008 Accepted: 6545 Description ‘Oh no, they’ve done it again’, crie…
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POJ 2533. Longest Ordered Subsequence
1. 题目 http://poj.org/problem?id=2533 Longest Ordered Subsequence Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 39666 Accepted: 17428 Description A numeric sequence of ai is ordered if a1 …
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URAL 1225. Flags
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1225 1225. Flags Time limit: 1.0 second Memory limit: 64 MB On the Day of the Flag of Russia a shop-owner decided to decorate the show-window of …
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URAL 1119. Metro
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1119 1119. Metro Time limit: 0.5 second Memory limit: 64 MB Many of SKB Kontur programmers like to get to work by Metro because the main office i…
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Floyd–Warshall算法简介
Floyd–Warshall算法用于寻找有向图中任意两点间的最短路径，即多源最短路径。该算法可用于带负权边的图，但不能用于带负权环的图。该算法需要比较每一组节点间所有的路径，时间复杂度为Θ(|V|^3)。 设有一个图G，有N个节点，编号为1,2,…,N。函数shortestPath(i, j, k) 返回当仅使用点集{1,2,…,k}为中间节点时，从节点i到达节点j…
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Tarjan算法简介
Tarjan算法由Robert Tarjan提出，可看作是Kosaraju算法的一个升级版，可用于获取有向图的强联通分量。该算法从任意起点开始进行DFS，每个节点只访问一次，其搜索树即为图的生成森林，图的强联通分量可以从森林的特定子树中恢复出来。 在使用栈进行DFS时，首先将任意节点入栈，然后递归地搜索与该节点相连的其他节点。某一节点u递归搜索结束时： 如果节点u和栈上节点u之前的其他节点…
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