SPF
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 9317   Accepted: 4218

Description

Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible.

Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate. 

Input

The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.

Output

For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist.

The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes.

Sample Input

1 2
5 4
3 1
3 2
3 4
3 5
0 1 2
2 3
3 4
4 5
5 1
0 1 2
2 3
3 4
4 6
6 3
2 5
5 1
0 0

Sample Output

Network #1
SPF node 3 leaves 2 subnets Network #2
No SPF nodes Network #3
SPF node 2 leaves 2 subnets
SPF node 3 leaves 2 subnets

Source

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
typedef long long LL;
#define MAXN 1009
#define N 100
/*
求割点的个数 和割点被去掉之后联通快的个数
//删除割点u产生的连通数目为:u所在的连通分量数目+与u所连接的割边的数目+1(边:fa->u)
*/
struct edge
{
//edge(int _t,int _next):to(_t),next(_next){}
int to, next;
};
edge E[MAXN * MAXN];
int index, dfn[MAXN], low[MAXN], s[MAXN], head[MAXN], cnt, Max, cas = , root;
bool flag;
void Init()
{
index = ;
Max = -;
cnt = ;
flag = false;
memset(head, -, sizeof(head));
memset(dfn, , sizeof(dfn));
memset(s, , sizeof(s));
memset(low, , sizeof(low));
}
void add_edge(int u, int v)
{
E[cnt].to = v;
E[cnt].next = head[u];
head[u] = cnt++;
Max = max(max(u, v), Max);
} void tarjan(int pre, int u)
{
int son = ;
low[u] = dfn[u] = ++index;
for (int i = head[u]; i != -; i = E[i].next)
{
int v = E[i].to;
if (!dfn[v])
{
tarjan(u, v);
son++;
low[u] = min(low[v], low[u]);
if ((u == root&&son > ) || (u != root&&low[v] >= dfn[u]))
{
flag = true;
s[u]++;
}
}
else if (v != pre)
low[u] = min(low[u], dfn[v]);
}
} int main()
{
int f, t;
while ()
{
Init();
while (scanf("%d",&f), f)
{
scanf("%d", &t);
add_edge(f, t);
add_edge(t, f);
}
if (Max == -)
break;
root = Max;
tarjan(-, Max);
printf("Network #%d\n",cas++);
if (!flag)
{
printf(" No SPF nodes\n");
}
else
{
for (int i = ; i <= Max; i++)
if (s[i] > )
printf(" SPF node %d leaves %d subnets\n", i, s[i] + );
}
printf("\n");
}
}

POJ 1523 SPF 割点 Tarjan的更多相关文章

  1. POJ 1523 SPF 割点与桥的推断算法-Tarjan

    题目链接: POJ1523 题意: 问一个连通的网络中有多少个关节点,这些关节点分别能把网络分成几部分 题解: Tarjan 算法模板题 顺序遍历整个图,能够得到一棵生成树: 树边:可理解为在DFS过 ...

  2. poj 1523 SPF(tarjan求割点)

    本文出自   http://blog.csdn.net/shuangde800 ------------------------------------------------------------ ...

  3. POJ 1523 SPF (割点,连通分量)

    题意:给出一个网络(不一定连通),求所有的割点,以及割点可以切分出多少个连通分量. 思路:很多种情况. (1)如果给的图已经不是连通图,直接“  No SPF nodes”. (2)求所有割点应该不难 ...

  4. POJ 1523 SPF tarjan求割点

                                                                   SPF Time Limit: 1000MS   Memory Limit ...

  5. Electricity POJ - 2117 + SPF POJ - 1523 去除割点后求强连通分量个数问题

    Electricity POJ - 2117 题目描述 Blackouts and Dark Nights (also known as ACM++) is a company that provid ...

  6. zoj 1119 / poj 1523 SPF (典型例题 求割点 Tarjan 算法)

    poj : http://poj.org/problem?id=1523 如果无向图中一个点 u 为割点 则u 或者是具有两个及以上子女的深度优先生成树的根,或者虽然不是一个根,但是它有一个子女 w, ...

  7. poj 1523 SPF(双连通分量割点模板)

    题目链接:http://poj.org/problem?id=1523 题意:给出无向图的若干条边,求割点以及各个删掉其中一个割点后将图分为几块. 题目分析:割点用tarjan算法求出来,对于每个割点 ...

  8. POJ 1523 SPF (去掉割点能形成联通块的个数)

    思路:使用tarjan算法求出割点,在枚举去掉每一个割点所能形成的联通块的个数. 注意:后来我看了下别的代码,发现我的枚举割点的方式是比较蠢的方式,我们完全可以在tarjan过程中把答案求出来,引入一 ...

  9. POJ 1523 SPF (无向图割点)

    <题目链接> 题目大意: 给你一个连通的无向图,问你其中割点的编号,并且输出删除该割点后,原图会被分成几个连通分量. 解题分析: Tarjan求割点模板题. #include <cs ...

随机推荐

  1. redis在linux的安装和开机启动(二)

    编译 安装 makefile已经存在 执行make 即可 make之后, 自动创建可运行的脚本文件, 不需要再执行 install了. 将脚本文件, 拷贝到指定位置, 就可以了. 手动创建目录, 需要 ...

  2. C#常量知识整理

    整数常量 整数常量可以是十进制.八进制或十六进制的常量.前缀指定基数:0x 或 0X 表示十六进制,0 表示八进制,没有前缀则表示十进制. 整数常量也可以有后缀,可以是 U 和 L 的组合,其中,U ...

  3. MyEclipse日志文件目录

    MyEclipse存放日志的目录为: <Workspace_Root>/.metadata/.log 在该目录下存放了日志的存档和当前日志,通过该日志可以查看MyEclipse的错误信息.

  4. service: no such service mysqld 与MySQL的开启,关闭和重启

    1.问题原因与解决办法 因为修改了MySQL临时文件的目录后,使用service mysqld restart重启MySQL出现如下错误: service: no such service mysql ...

  5. 297 Serialize and Deserialize Binary Tree 二叉树的序列化与反序列化

    序列化是将一个数据结构或者对象转换为连续的比特位的操作,进而可以将转换后的数据存储在一个文件或者内存中,同时也可以通过网络传输到另一个计算机环境,采取相反方式重构得到原数据.请设计一个算法来实现二叉树 ...

  6. JAVA使用Ldap操作AD域

    项目上遇到的需要在集成 操作域用户的信息的功能,第一次接触ad域,因为不了解而且网上其他介绍不明确,比较费时,这里记录下. 说明: (1). 特别注意:Java操作查询域用户信息获取到的数据和域管理员 ...

  7. spring中配置数据源

    spring中配置数据源的几种常见方式: #mysql 数据库配置(jdbc.properties) jdbc.driverClassName=com.mysql.jdbc.Driver jdbc.u ...

  8. JAVA 学习笔记 - 基础语法1

    1. 类的定义 public class  与  class的区别 一个java文件只有一个public class, 而且类名必须与文件名一致. 一个java文件可以有多个class定义,javac ...

  9. VS2015 安装包缺失(联网安装失败)问题解决

    Win7 x86 测试可行 *  如果前面有尝试过安装不成功, 一定要用卸载程序删除已安装的部分,否则会出乱子. 1. 或者是用虚拟光驱加载ISO, 或者是解压到硬盘上, 都没有关系. 2. 用管理员 ...

  10. vue项目国际化实现 vue-i18n使用详细教程

    1.安装vue-i18n: npm i vue-i18n -S 当然你也可以这样: <script src="https://unpkg.com/vue/dist/vue.js&quo ...