You've got a undirected graph G, consisting of n nodes. We will consider the nodes of the graph indexed by integers from 1 to n. We know that each node of graph G is connected by edges with at least k other nodes of this graph. Your task is to find i…
答案=总数-无0-无1-无2+无01+无02+无12-无012 直接详细讲无0和无2 无0为 01和11,无2为01和00,显然二者方案数相同,以下考虑无0 考虑折半搜索,后半段搜索,二进制点权0的位置,保证后半段构成的无0边的基础上 可以得出一个S集合,表示集合内的点随意选择,不在集合内的点只能为1 再搜索前半段,同后半段搜索,得出一个集合T里面的点权为0,则T的贡献为满足\(T\subseteq S\)的S的个数,这个子集和一下即可…
描述 Given a directed graph containing n vertice (numbered from 1 to n) and m edges. Can you tell us how many different Hamiltonian Cycles are there in this graph? A Hamiltonian Cycle is a cycle that starts from some vertex, visits each vertex (except…
A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes u and v (u ≠ v) exists either an edge going from u to v, or an edge from v to u. You are give…
Source: Connected Brain Figure above: Bullmore E, Sporns O. Complex brain networks: graph theoretical analysis of structural and functional systems.[J]. Nature Reviews Neuroscience, 2009, 10(3):186-198. Graph measures A graph G consisting of a set of…
Graph cuts是一种十分有用和流行的能量优化算法,在计算机视觉领域普遍应用于前背景分割(Image segmentation).立体视觉(stereo vision).抠图(Image matting)等. 此类方法把图像分割问题与图的最小割(min cut)问题相关联.首先用一个无向图G=<V,E>表示要分割的图像,V和E分别是顶点(vertex)和边(edge)的集合.此处的Graph和普通的Graph稍有不同.普通的图由顶点和边构成,如果边的有方向的,这样的图被则称为有向图,否则为…
Description P. T. Tigris is a student currently studying graph theory. One day, when he was studying hard, GS appeared around the corner shyly and came up with a problem: Given a graph with n nodes and m undirected weighted edges, every node having o…
题目传送门 暴力搜索 看到这道题的第一反应就是直接上$bfs$啦,也没有想到什么更加优秀的算法. 然后就是$15$分钟打了$70$分,有点震惊,纯暴力诶,这么多白给分嘛,太划算了,这可是$D2T3$诶. #include<cstdio> #include<algorithm> #include<vector> #include<cstring> #include<queue> #include<map> #include<ios…
题目链接:http://codeforces.com/problemset/problem/263/D 思路:一遍dfs即可,dp[u]表示当前遍历到节点u的长度,对于节点u的邻接点v,如果v没有被访问过,则继续访问,否则计算dp[u] - dp[v] + 1是否大于等于K + 1,如果是,就说明找到了这样一个符合要求的环,然后将回退,回退的时候将环上的节点标记即可,否则,就继续找. #include <iostream> #include <cstdio> #include &l…
题意:给出n个点的度数列 上限(实际点可以小于该度数列)问可以构造简单路最大长度是多少(n个点要连通 不能有平行边.重边) 思路:直接构造一条长链  先把度数为1的点 和度数大于1的点分开  先把度数大于1的点连在一起 然后把度数为1的点连在两边可以涨最多2的长度(如果有大于等于2的度数为1的点) 随后就涨不了长度了,还要把度数为1的点接在链上 判断是否可以构成这样一颗树 (在while里面少写了度数-- WA了两次7 QAQ) #include<bits/stdc++.h> #define…