judge loop in undirected graph
一 深度优先遍历,参考前面DFS(white and gray and black)
二 根据定点以及边数目进行判断
如果m(edge)大于n(vertex),那么肯定存在环
算法如下:
1 删除所有入度小于等于1的顶点, 并且将和这些顶点相关的顶点入度减1
2 将入度变为1的顶点全部删除,重复上述动作,如果最后还有顶点那么图中存在环
具体代码如下:
#include <iostream>
using namespace std; #define MAX_VERTEX_NUM 128
enum color{WHITE, GRAY = 1, BLACK};
bool M[MAX_VERTEX_NUM][MAX_VERTEX_NUM];
int colour[MAX_VERTEX_NUM];
int dfsNum[MAX_VERTEX_NUM], num;
int indegree[MAX_VERTEX_NUM];
int vexnum, edgenum; void init_graph(){
cout<<"enter vertex number:"<<endl;
cin>>vexnum;
cout<<"enter edge number:"<<endl;
cin>>edgenum; int i, j;
while(edgenum){
cout<<"add new edge:"<<endl;
cin>>i>>j;
M[i - 1][j - 1] = true;
//initialize in vertex degree
indegree[i - 1]++;
indegree[j - 1]++;
edgenum--;
}
}
/*
void dfs(int u, int p){
colour[u] = GRAY;
dfsNum[u] = num++;
for( int v = 0; v < vexnum; v++){
if(M[u][v] && v != p){
if(colour[v] == WHITE) dfs(v, u);
else if(colour[v] == GRAY)
cout<<"back edge between"<<u + 1<<" and"<<v + 1<<endl;
else if(colour[v] == BLACK)
cout<<"cross edge between"<<u + 1<<" and"<<v + 1<<endl;;
}
}
colour[u] = BLACK;
}
void print_dfs_num(){
for(int v = 0; v < vexnum; v++)
cout<<dfsNum[v]<<" ";
}
*/ void LoopJudge(){
bool loop = false; int twice = 2;
int k, i, j;
cout<<"line: "<<__LINE__<<endl;
for( k = twice; k > 0; k--){
cout<<"line: "<<__LINE__<<"k: "<<k<<endl;
for( i = 0; i < vexnum; i++){
cout<<"line: "<<__LINE__<<"i: "<<i<<endl;
if(indegree[i] <= 1){
indegree[i] = 0; //delete vertex in degree equal one
for( j = 0; j < vexnum; j++){
cout<<"line: "<<__LINE__<<"j: "<<j<<endl;
if(M[i][j]){
M[i][j] = false;
M[j][i] = false;
indegree[j]--;
}//if(M[i][j])
}//for(int j = 0; j < vexnum; j++)
}//if(indegree[i] <= 1)
}//for(int i = 0; i < vexnum; i++)
} for( k = 0; k < vexnum; k++){
if(indegree[k] != 0){
loop = true;
}
} if(loop)
cout<<"There is loop in undirected graph!"<<endl;
else
cout<<"There is no loop in undirected graph!"<<endl;
} int main()
{
init_graph();
//dfs(0, -1);
//print_dfs_num();
LoopJudge(); int ch;
cin>>ch;
return 0;
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
judge loop in undirected graph的更多相关文章
- Judge loop in directed graph
1 深度优先方法 首先需要更改矩阵初始化函数init_graph() 然后我们需要初始化vist标记数组 深度优先访问图,然后根据是否存在back edge判断是否存在环路 算法如下: #includ ...
- [LeetCode] Number of Connected Components in an Undirected Graph 无向图中的连通区域的个数
Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), ...
- LeetCode Number of Connected Components in an Undirected Graph
原题链接在这里:https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/ 题目: Giv ...
- Leetcode: Graph Valid Tree && Summary: Detect cycle in undirected graph
Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), ...
- lintcode:Find the Connected Component in the Undirected Graph 找出无向图汇总的相连要素
题目: 找出无向图汇总的相连要素 请找出无向图中相连要素的个数. 图中的每个节点包含其邻居的 1 个标签和 1 个列表.(一个无向图的相连节点(或节点)是一个子图,其中任意两个顶点通过路径相连,且不与 ...
- [Locked] Number of Connected Components in an Undirected Graph
Number of Connected Components in an Undirected Graph Given n nodes labeled from 0 to n - 1 and a li ...
- [Swift]LeetCode323. 无向图中的连通区域的个数 $ Number of Connected Components in an Undirected Graph
Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), ...
- [LintCode] Find the Connected Component in the Undirected Graph
Find the Connected Component in the Undirected Graph Find the number connected component in the undi ...
- 323. Number of Connected Components in an Undirected Graph按照线段添加的并查集
[抄题]: Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of n ...
随机推荐
- Netty那点事: 概述, Netty中的buffer, Channel与Pipeline
Netty那点事(一)概述 Netty和Mina是Java世界非常知名的通讯框架.它们都出自同一个作者,Mina诞生略早,属于Apache基金会,而Netty开始在Jboss名下,后来出来自立门户ne ...
- CCNP路由实验(1) -- EIGRP
EIGRP(Enhanced Interior Gateway Routing Protocol,增强型内部网关路由协议)是Cisco公司开发的一个平衡混合型路由协议,它融合了距离向量和链路状态两种路 ...
- Finding the Longest Palindromic Substring in Linear Time
Finding the Longest Palindromic Substring in Linear Time Finding the Longest Palindromic Substring i ...
- C#多线程及GDI(Day 23)
又来到了总结知识的时间了,今天又学了一些新的知识,是多线程和GDI的一些运用. 理论: 在学习多线程之前,首先要了解一下什么是进程? 进程:(关键字Process)进程是一个具有一定独立功能的程 ...
- Latex调整行距
修改行间距的方法: \usepackage{setspace}%使用间距宏包 \begin{document} \begin{spacing}{2.0}%%行间距变为double-space 双倍行距 ...
- Best practice for Invoke other scripts or exe in PowerShell
Recently, I find I used many different type method to invoke other scripts or exe in PowerShell. May ...
- 高质量程序设计指南C/C++语言——C++/C程序设计入门(4)
*switch结构的break语句只是一个“jmp”指令,其作用就是跳到switch结构的结尾处 *标准C++/C语言提供3种循环:do/while.while和for,它们都在条件表达式为TRUE( ...
- 带你一起Piu Piu Piu~
单刀直入,今天要讲的是自己写的一个WPF动画例子.我们在看下最终效果~ 最近在重看WPF编程宝典2010,在练习第15章动画性能例子时有了些想法.原始例子如下: 原始例子(打包了整个15章的) 它是 ...
- Spring Cache使用详解
Spring Cache Spring Cache使用方法与Spring对事务管理的配置相似.Spring Cache的核心就是对某个方法进行缓存,其实质就是缓存该方法的返回结果,并把方法参数和结果用 ...
- poj 2375 Cow Ski Area bfs
这个题目用tarjan找联通块,缩点,然后统计出入度为0的点理论上是可行的,但问题是会暴栈.考虑到这个题目的特殊性,可以直接用一次bfs找到数字相同且联通的块,这就是一个联通块,然后缩点,统计出入度即 ...