1 // 把一个图的所有结点排序,使得每一条有向边(u,v)对应的u都排在v的前面. 2 // 在图论中,这个问题称为拓扑排序.(toposort) 3 // 不难发现:如果图中存在有向环,则不存在拓扑排序,反之则存在. 4 // 不包含有向环的有向图称为有向无环图(DAG). 5 // 可以借助DFS完成拓扑排序:在访问完一个结点之后把它加到当前拓扑序的首部. 6 7 int c[maxn]; 8 int topo[maxn],t; 9 bool dfs(int u) 10 { 11 c[u]…

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 Krypton Factor  You have been employed by the organisers of a Super Krypton Factor Contest in which contestants have very high mental and physical abilities. In one section of the contest the contestants are tested on their ability to recall a seque…

After the last war devastated your country, you - as the king of the land of Ardenia - decided it washigh time to improve the defense of your capital city. A part of your fortication is a line of magetowers, starting near the city and continuing to t…

Emily the entrepreneur has a cool business idea: packaging and selling snowakes. She has devised amachine that captures snowakes as they fall, and serializes them into a stream of snowakes that ow,one by one, into a package. Once the package is full,…

Oil Deposits  The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. I…

题意: 有n种正放形,每种正方形的数量可视为无限多.已知边与边之间的结合规则,而且正方形可以任意旋转和反转,问这n中正方形是否可以拼成无限大的图案. 分析: 首先因为可以旋转和反转,所以可以保证在拼接的过程中正方形不会自交. 把边的标号看成点,将正方形的边界A+变成B+可以看做是一条边.比如说,一个正方形中有A-和B+两条边,则A-与其他正方形中A+结合后,结合前边界为A-,结合后变为B+. 这样就得到图中的一条有向边A+ → B+ 如果能在图中找到一个环,则可以无限循环拼接正方形. #incl…

拓扑排序,以边上标号为点,正方形为边,拓扑图中存在有向环时unbounded,否则bounded: 注意:仔细处理输入:   遍历一个点时,下一次遍历拼上的下一个方形边:即假设遍历到 A+ 时,下次从 A- 开始遍历: #include <iostream> #include <cstdio> #include <cstring> using namespace std; int n; ][]; ]; int dfs (int u){ visit[u]=-; ) u--…

题目链接: https://cn.vjudge.net/problem/UVA-1572 Automatic Chemical Manufacturing is experimenting with a process called self-assembly. In this process, molecules with natural affinity for each other are mixed together in a solution and allowed to sponta…