zoj3732&& hdu4797 Graph Reconstruction】的更多相关文章

Graph Reconstruction Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Let there be a simple graph with N vertices but we just know the degree of each vertex. Is it possible to reconstruct the graph only by these information? A sim…
Graph Reconstruction Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Let there be a simple graph with N vertices but we just know the degree of each vertex. Is it possible to reconstruct the graph only by these information? A sim…
C. Graph Reconstruction Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/329/problem/C Description I have an undirected graph consisting of n nodes, numbered 1 through n. Each node has at most two incident edges. For each pa…
题目链接 Graph Reconstruction 题意 给你无向图每个点的度数, 问是否存在唯一解, 存在输出唯一解, 多解输出两个, 无解输出IMPOSSIBLE 思路 这里用到了 Havel-Hakimi定理, 实际上就是按照度数构图的一种贪心策略. 这样能判断出来一个解或无解, 多解的情况, 只要在比较构图时最后面两个点的当前度数一样, 那么选其一都是可行的, 但是选其一的情况有保证了另外一个肯定没有连上, 所以能找到其它解. 代码 #include <bits/stdc++.h> #…
分析: 给定一个非负整数序列{dn},若存在一个无向图使得图中各点的度与此序列一一对应,则称此序列可图化. 进一步,若图为简单图,则称此序列可简单图化 (来自百度百科) 可简单图化的判定可以用Havel-Hakimi定理,然后简述 Havel-Hakimi定理 Havel-Hakimi定理的过程: 1,按度数排序. 2,选取度数最大的点,如果该点度数为0,结束,有解 3,每次选一个度数最大的点,然后将后面的点的度数依次减1,表示该顶点和相应的顶点有边相连, 如果有点的度数减到负数,结束,无解.…
I have an undirected graph consisting of n nodes, numbered 1 through n. Each node has at most two incident edges. For each pair of nodes, there is at most an edge connecting them. No edge connects a node to itself. I would like to create a new graph…
C. Graph Reconstruction time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output I have an undirected graph consisting of n nodes, numbered 1 through n. Each node has at most two incident edges. Fo…
论文信息 论文标题:Structural Entropy Guided Graph Hierarchical Pooling论文作者:Junran Wu, Xueyuan Chen, Ke Xu, Shangzhe Li论文来源:2022,ICLR论文地址:download 论文代码:download 1 Introduction 之前池化的方法大多是 固定比例的池化 或者 基于逐步池化设计的.本文针对上述问题,设计了 SEP 模型,具体来说,在不分配特定于层的压缩配额的情况下,设计了一种全局优…
题目链接: http://codeforces.com/contest/330 A. Cakeminator 题目意思: 给一个r*c的矩阵方格,有些位置有S,如果某一行和一列都不含标记为S的方格,则可以把该行所有方格都收掉,问最多能收多少个方格,方格可以收多次,多次收的方格计数一次. 解题思路: 暴力 代码: #include<iostream> #include<cmath> #include<cstdio> #include<cstdlib> #inc…
第一次就去拉了点思维很神奇的CF题目 2018省赛赛第一次训练 # Origin Title     A CodeForces 607A Chain Reaction     B CodeForces 385C Bear and Prime Numbers     C CodeForces 670D2 Magic Powder - 2     D CodeForces 360B Levko and Array     E CodeForces 68B Energy exchange     F…