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Question: Judge the key edge in the maximum matching of a bipartite graph, that is, remove these edges and you will not be able to get the maximum matching. As the question shows a 100*100 graph, the enumeration + Hungary complexity is 10 ^ 8.

Finally, I ended all my 20XX questions. I had never dared to start with this question before. Now I have learned the recursive equation of matrix optimization, and this question is coming soon... X [I] indicates that the number of I characters A

Simple TarjanAlgorithm. In fact, the Tarjan algorithm first finds the root of a strongly connected component tree, and then it is easy to find the entire strongly connected component. The key is to use the DFS tree.   Maze castle Time Limit: 200

Basic Joseph's ring Remember the recursive formula of Joseph's ring. Assume that it is the K-person from 0 to n-1 and from 0 F [1] = 0; F [I] = (F [I-1] + k + 1) % I (I> 1)   Eeny meeny moo Time limit:

A simple bipartite graph. Idea: For each vertex label that is not Ponds, split each vertex into two during graph creation, and perform the maximum matching between the two parts. The specific method of edge creation is that each vertex is labeled,

First, determine whether it is a bipartite graph. DFS will dye each vertex. If there is no conflict, it means a bipartite graph. Then Hungary goes through water directly...   The accomodation of Students time limit: 5000/1000 MS (Java/others)

Evaluate how many characters must be added to a string before it can be converted into a return text. Idea: Find the largest public subsequence of this string and its inverse string. In fact, this sequence is actually the largest number of input

Optimal Match of KM. KmAlgorithmIs to give each vertex a label (calledTop logo. Set vertex XI to a [I], vertex Yi to B [I], and edge weight between vertex XI and YJ to W [I, j]. At any time in the Algorithm Execution Process, for any side (I, j),

It is implemented using the semi-segmentation method. after dividing a heap into two heaps and recursively obtaining the minimum distance between the two heaps, the key is to determine whether the distance between the two heaps is smaller, then we

Recent common ancestor .. It indicates that it has been stuck in a single location for a long time .. Thinking is still a little worse. to record the distance from a tree, you only need to record the distance to the root node. If the distance

The key to a network stream with the upper and lower bounds is how to reflect the lower bounds. Therefore, you must first determine whether the lower bounds are feasible. And then solve the maximum stream. This question does not provide the source

# Include # Include String . H> # Include Using Namespace STD; # Define N 250 # Define M 20100 # Define INF 0x3fffffff Struct Node { Int From , To, next, W;} edge [m]; Int N, m; Int CNT, pre [N]; Int S1, NN, T; Int LV [N], gap [N];

KM simple question, but pay attention to some things. 1. the 500*500 Point Graph mentioned in the question cannot be processed based on the complexity of the km. However, because the data of this question is not enough, it is not a matter of

This question can be obtained by performing row or column transformations based on a square matrix with a diagonal line of 1. If you want to do this, you can know that each row selects only one column. You need to select all the columns, and you

The key to this question is to convert A + B + C = D to A + B = D-C, and then sort all the situations of A + B, then, enumerate D and C in the ascending order, and search for D-C in the second place.   Sumsets

At first, I thought that this question was a Bipartite Graph Match. Then I used Hungary to do it, TLE, and HK to do it, TLE... It is found that the network stream is used, and 100000 vertices need to be compressed, because the possible conditions

Stable sorting is a sorting method that does not change after the same elements are sorted. Bubble is a stable sorting. A faster merge is also a stable sorting.  Stable sorting Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/

Greedy: first use dfs to record all vertices from node 1, and then reverse greedy. This ensures that each time the child node is judged and then the parent node is judged. Then, when judging every two adjacent points (one must be taken), we select

Find the number of LCM, you can know that from the first one, find the minimum public multiple for each two, and then get the answer.Least Common Multiple Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total

The first DLX question is a template question. After reading the paper for a long time, I learned that the first write of the two-way cross-linked list is still a bit tangled. But be patient. Search Optimization, I feel that this ultra-efficient