PTA Strongly Connected Components
Write a program to find the strongly connected components in a digraph.
Format of functions:
void StronglyConnectedComponents( Graph G, void (*visit)(Vertex V) );
where Graph is defined as the following:
typedef struct VNode *PtrToVNode;
struct VNode {
Vertex Vert;
PtrToVNode Next;
};
typedef struct GNode *Graph;
struct GNode {
int NumOfVertices;
int NumOfEdges;
PtrToVNode *Array;
};
Here void (*visit)(Vertex V) is a function parameter that is passed into StronglyConnectedComponents to handle (print with a certain format) each vertex that is visited. The function StronglyConnectedComponents is supposed to print a return after each component is found.
Sample program of judge:
#include <stdio.h>
#include <stdlib.h>
#define MaxVertices 10 /* maximum number of vertices */
typedef int Vertex; /* vertices are numbered from 0 to MaxVertices-1 */
typedef struct VNode *PtrToVNode;
struct VNode {
Vertex Vert;
PtrToVNode Next;
};
typedef struct GNode *Graph;
struct GNode {
int NumOfVertices;
int NumOfEdges;
PtrToVNode *Array;
};
Graph ReadG(); /* details omitted */
void PrintV( Vertex V )
{
printf("%d ", V);
}
void StronglyConnectedComponents( Graph G, void (*visit)(Vertex V) );
int main()
{
Graph G = ReadG();
StronglyConnectedComponents( G, PrintV );
return 0;
}
/* Your function will be put here */
Sample Input (for the graph shown in the figure):
4 5
0 1
1 2
2 0
3 1
3 2
Sample Output:
3
1 2 0
Note: The output order does not matter. That is, a solution like
0 1 2
3
is also considered correct.
这题目就是直接照搬Tarjan算法实现就好了,Tarjan算法在《算法导论》上第22章有,但是我看了以后并没有明白Tarjan算法的过程orz,最后还是看blog看懂的,所以推荐一个讲Tarjan算法讲的很好的blog:http://blog.csdn.net/acmmmm/article/details/16361033 还有Tarjan算法实现的具体代码:http://blog.csdn.net/acmmmm/article/details/9963693 都是一个ACM大佬写的,我就是看这两个的……其实我也看了很久才看懂Tarjan算法是干啥的……毕竟上课从来不听不知道老师讲的方法是怎么样的……
当然只要理解了Tarjan算法,这题目就相当easy了。
补充:还看到一个英文的讲Tarjan的地方,讲的很全面,在geeksforgeeks上http://www.geeksforgeeks.org/tarjan-algorithm-find-strongly-connected-components/
就是打开可能会有点慢,但是不需要FQ。
直接放代码吧:
//
// main.c
// Strongly Connected Components
//
// Created by 余南龙 on 2016/12/6.
// Copyright © 2016年 余南龙. All rights reserved.
//
int dfn[MaxVertices], low[MaxVertices], stack[MaxVertices], top, t, in_stack[MaxVertices];
int min(int a, int b){
if(a < b){
return a;
}
else{
return b;
}
}
void Tarjan(Graph G, int v){
PtrToVNode node = G->Array[v];
int son, tmp;
dfn[v] = low[v] = ++t;
stack[++top] = v;
in_stack[v] = ;
while(NULL != node){
son = node->Vert;
== dfn[son]){
Tarjan(G, son);
low[v] = min(low[son], low[v]);
}
== in_stack[son]){
low[v] = min(low[v], dfn[son]);
}
node = node->Next;
}
if(dfn[v] == low[v]){
do{
tmp = stack[top--];
printf("%d ", tmp);
in_stack[tmp] = ;
}while(tmp != v);
printf("\n");
}
}
void StronglyConnectedComponents( Graph G, void (*visit)(Vertex V) ){
int i;
; i < MaxVertices; i++){
dfn[i] = -;
low[i] = in_stack[i] = ;
}
top = -;
t = ;
; i < G->NumOfVertices; i++){
== dfn[i]){
Tarjan(G, i);
}
}
}
PTA Strongly Connected Components的更多相关文章
- Strongly connected components
拓扑排列可以指明除了循环以外的所有指向,当反过来还有路可以走的话,说明有刚刚没算的循环路线,所以反过来能形成的所有树都是循环
- algorithm@ Strongly Connected Component
Strongly Connected Components A directed graph is strongly connected if there is a path between all ...
- [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 ...
- [Redux] Using withRouter() to Inject the Params into Connected Components
We will learn how to use withRouter() to inject params provided by React Router into connected compo ...
- [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 ...
- cf475B Strongly Connected City
B. Strongly Connected City time limit per test 2 seconds memory limit per test 256 megabytes input s ...
- Strongly connected(hdu4635(强连通分量))
/* http://acm.hdu.edu.cn/showproblem.php?pid=4635 Strongly connected Time Limit: 2000/1000 MS (Java/ ...
- [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), ...
随机推荐
- TODO的使用
在vs2012中使用TODO添加注释
- oracle 单列索引 多列索引的性能测试
清除oralce 缓存:alter system flush buffer_cache; 环境:oracle 10g . 400万条数据,频率5分钟一条 1.应用场景: 找出所有站点的最新一条数据. ...
- SVN Cornerstone 报错信息 xcodeproj cannot be opened because the project file cannot be parsed.
svn点击update 之后,打开xcode工程文件,会出现 xxx..xcodeproj cannot be opened becausethe project file cannot be p ...
- 学习总结——Selenium元素定位
读一本好书,不能读读就算了,做一下总结,变成自己的,以备查阅. 1. driver.findElement(By.id(<element ID>)) ID是独一无二的,使用 ...
- 13,SFDC 管理员篇 - 移动客户端
1, 自定义导航 设置导航显示内容 Setup | Mobile Administration | Salesforce Navigation 1, 可以添加和删除在mobile中显示的内容 ...
- 27. Oracle 10g下emctl start dbconsole 报错:OC4J Configuration issue 问题解决
(dbconsole配置好外面的sqlplus才能连的上服务器上的数据库) 采取重新配置emctl 的方法来解决 [oracle@guohuias3 oracle]$ emca -config dbc ...
- tomcat manager配置
在tomcat-user.xml里面配置 <tomcat-users xmlns="http://tomcat.apache.org/xml" xmlns:xsi=" ...
- XidianOJ 1041: Franky的游戏O
题目描述 Franky是super的人造人,来到了n*m的棋盘世界玩冒险游戏. n×m的棋盘由n行每行m个方格组成,左上角的方格坐标是(0,0),右下角的方格坐标是(n-1,m-1). 每次游戏时,他 ...
- ubuntu 配置nginx+php+mysql 遇到的一些问题
/* 公司内网打算配置一台ubuntu为主机的测试服务器.刚好手头有一个昂达的主机,装的windows 声音又大,还不如直接装ubuntu .声音又小,还占用资源少. */ 刚开始安装php5 结果提 ...
- 转:python中对list去重的多种方法
对一个list中的新闻id进行去重,去重之后要保证顺序不变. 直观方法 最简单的思路就是: ids = [1,2,3,3,4,2,3,4,5,6,1] news_ids = [] for id in ...