Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 8254   Accepted: 3772

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

点双连通分量模板题

 /*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=;
struct edge{
int v,nxt;
}e[mxn<<];
int hd[mxn],mct=;
void add_edge(int u,int v){
e[++mct].v=v;e[mct].nxt=hd[u];hd[u]=mct;
return;
}
int n;
int low[mxn],dfn[mxn],dtime=;
int num[mxn];
int st[mxn],top=;
bool cut[mxn];
void DFS(int u,int fa){
low[u]=dfn[u]=++dtime;
st[++top]=u;
int i,j;
for(i=hd[u];i;i=e[i].nxt){
int v=e[i].v;
if(v==fa)continue;
if(!dfn[v]){
DFS(v,u);
if(low[v]>=dfn[u]){
cut[u]=;
int s=;
do{
s=st[top--];
++num[s];
}while(u!=s);
top++;
}
// num[u]++;
low[u]=min(low[u],low[v]);
}
else low[u]=min(dfn[v],low[u]);
}
return;
}
bool have[mxn];
int main(){
int cas=;
while(){
memset(hd,,sizeof hd);
memset(cut,,sizeof cut);
memset(low,,sizeof low);
memset(dfn,,sizeof dfn);
memset(have,,sizeof have);
memset(num,,sizeof num);
mct=dtime=top=;
int i,j,u,v;
bool flag=;
while(scanf("%d",&u) && u){
scanf("%d",&v);
have[u]=have[v]=;
add_edge(u,v);
add_edge(v,u);
flag=;
}
if(!flag)break;
for(i=;i<=;i++)
if(have[i] && !dfn[i])DFS(i,);
printf("Network #%d\n",++cas);
flag=;
for(i=;i<=;i++){
if(num[i]>=){
flag=;
printf(" SPF node %d leaves %d subnets\n",i,num[i]);
}
}
if(!flag)printf(" No SPF nodes\n");
printf("\n");
}
return ;
}

POJ1523 SPF的更多相关文章

  1. POJ1523 SPF[无向图割点]

    SPF Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8139   Accepted: 3723 Description C ...

  2. POJ1523 SPF 单点故障

    POJ1523 题意很简单,求删除割点后原先割点所在的无向连通图被分成了几个连通部分(原题说prevent at least one pair of available nodes from bein ...

  3. POJ1523 SPF(割点模板)

    题目求一个无向图的所有割点,并输出删除这些割点后形成几个连通分量.用Tarjan算法: 一遍DFS,构造出一颗深度优先生成树,在原无向图中边分成了两种:树边(生成树上的边)和反祖边(非生成树上的边). ...

  4. POJ1523:SPF(无向连通图求割点)

    题目:http://poj.org/problem?id=1523 题目解析: 注意题目输入输入,防止PE,题目就是求割点,并问割点将这个连通图分成了几个子图,算是模版题吧. #include < ...

  5. poj图论解题报告索引

    最短路径: poj1125 - Stockbroker Grapevine(多源最短路径,floyd) poj1502 - MPI Maelstrom(单源最短路径,dijkstra,bellman- ...

  6. Tarjan求割点和桥

    by szTom 前置知识 邻接表存储及遍历图 tarjan求强连通分量 割点 割点的定义 在一个无向图中,如果有一个顶点集合,删除这个顶点集合以及这个集合中所有顶点相关联的边以后,图的连通分量增多, ...

  7. POJ1523:SPF——题解

    http://poj.org/problem?id=1523 这题明显就是求割点然后求割完之后的强连通分量的个数. 割点都会求,怎么求割完的分量个数呢? 我们可以通过万能的并查集啊!(具体做法看代码吧 ...

  8. POJ 1523 SPF(求割点)

    题目链接 题意 : 找出图中所有的割点,然后输出删掉他们之后还剩多少个连通分量. 思路 : v与u邻接,要么v是u的孩子,要么u是v的祖先,(u,v)构成一条回边. //poj1523 #includ ...

  9. 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 ...

随机推荐

  1. SpringMVC系列之基本配置

    一.概述 Spring Web MVC是一种基于Java的实现了Web MVC设计模式的请求驱动类型的轻量级Web框架,即使用了MVC架构模式的思想,将web层进行职责解耦,基于请求驱动指的就是使用请 ...

  2. java 13-6 Char的包装类Character

    1.Character 类在对象中包装一个基本类型 char 的值 此外,该类提供了几种方法,以确定字符的类别(小写字母,数字,等等),并将字符从大写转换成小写,反之亦然 构造方法: Characte ...

  3. 动态调用webservice,不需要添加Web References

    using System; using System.Collections.Generic; using System.Web; using System.Net; using System.IO; ...

  4. TeeChart控件的安装与常用 功能设置

    TeeChart控件的安装 TeeChart 7.0 With Source在Delphi 7.0中的安装 一.删除Delphi7自带TeeChart  1.Component -> insta ...

  5. 小程序基础09:视图层之WXML

    1.WXML WXML是框架设计的一套标签语言,结合基础组件,事件系统,可以构建出页面的结构. 用以下的例子来看看WXML有什么能力: 1.1数据绑定 <view>{{age}}</ ...

  6. 使用ajax跨域withCredentials的作用

    默认情况下,跨源请求不提供凭据(cookie.HTTP认证及客户端SSL证明等).通过将withCredentials属性设置为true,可以指定某个请求应该发送凭据.如果服务器接收带凭据的请求,会用 ...

  7. Android Studio Gradle编译项目报错

    Gradle project sync failed Android Studio每次更新版本都会更新Gradle这个插件,但由于长城的问题每次更新都是失败,又是停止在Refreshing Gradl ...

  8. 关闭Outlook时最小化 dll

    用Outlook时最让我感觉不爽的就是不小心点了关闭按钮就会把Outlook关闭. 我们用软件的时候都希望软件有一个关闭时最小化功能,更希望Outlook也有这个功能 但让我很失望的是把设置里看了一个 ...

  9. 如何使用impress.js做一个网页版本的PPT

    blockquote{font-size: 18px;line-height:1.5;margin:0;}line-height: 1.5; 要做一个网站制作规范培训,之前村长做过一次培训,但是后来一 ...

  10. HTC Vive 体验的折腾经历

    HTC Vive 是个什么东西, 想必我就不用介绍了, 不知道自己百度吧 HTC Vive发布已经有一段时间了, 一直很纠结买还是不买, 这玩意太贵(官网6888),买了还不能直接用, 还要配太高性能 ...