https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=251

 Network 

A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N. No two places have the same number. The lines are bidirectional and always connect together two places and in each place the lines end in a telephone exchange. There is one telephone exchange in each place. From each place it is possible to reach through lines every other place, however it need not be a direct connection, it can go through several exchanges. From time to time the power supply fails at a place and then the exchange does not operate. The officials from TLC realized that in such a case it can happen that besides the fact that the place with the failure is unreachable, this can also cause that some other places cannot connect to each other. In such a case we will say the place (where the failure occured) is critical. Now the officials are trying to write a program for finding the number of all such critical places. Help them.

Input

The input file consists of several blocks of lines. Each block describes one network. In the first line of each block there is the number of places N < 100. Each of the next at most N lines contains the number of a place followed by the numbers of some places to which there is a direct line from this place. These at mostN lines completely describe the network, i.e., each direct connection of two places in the network is contained at least in one row. All numbers in one line are separated by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0.

Output

The output contains for each block except the last in the input file one line containing the number of critical places.

Sample Input

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

Sample Output

1
2
 
题目大意:
给你一个无向图,求其中割点的个数目。
输入数据
第一行一个 n 代表有 n 个点
接下来有多行,一直到读入一个 0,算整个地图的读入结束,再读入一个0,文件数据结束。
每行有第一个数字a,代表接下来的数字都 和 a 相连。 
 
 
割点:无向连通图中,如果删除某点后,图变成不连通了,则称该点为割点。
这里割点 和 桥 都是无向图里的概念,大家在这里不要混淆了。
 
求割点
一个顶点u是割点,当且仅当满足(1)或(2)
(1) u为树根,且u有多于一个子树。  
即代码中rootson >1
(2) u不为树根,且满足存在(u,v)为树枝边(或称 父子边,即u为v在搜索树中的父亲),使得 dfn(u)<=low(v)。(也就是说 V 没办法绕过 u 点到达比 u dfn要小的点)   即代码中  if(dfn[v] <= low[i])   Cut[i] = true;
注:这里所说的树是指,DFS下的搜索树。
 
求割点 Tarjan里 low  和  dfn
dfn[u]定义和前面类似,但是low[u]定义为u
或者u的子树中能够通过非父子边追溯到的
最早的节点的DFS开始时间
在Tarjan算法求割点我们要加一个数组 f[u], 判断两者是否是父子边
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#define N 110
#define min(a, b)(a < b ? a : b) using namespace std; vector<vector<int> >G;
int low[N], dfn[N], f[N];
int n, Time, num;
bool Cut[N]; void Init()
{
G.clear();
G.resize(n + );
memset(low, , sizeof(low));//最先到达该点的时间
memset(dfn, , sizeof(dfn));//该点能到达之前点的最早时间
memset(f, , sizeof(f));//保存一个点的父节点
memset(Cut, false, sizeof(Cut));//判断该是否为割点
Time = num = ;
} void Tarjan(int u, int fa)
{
int len, v, i;
low[u] = dfn[u] = ++Time;
f[u] = fa;
len = G[u].size();
for(i = ; i < len ; i++)
{
v = G[u][i];
if(!dfn[v])
{
Tarjan(v, u);
low[u] = min(low[u], low[v]);
}
else if(fa != v)
low[u] = min(low[u], dfn[v]);
}
} void Solve()
{
int rootson = , i, v;
Tarjan(, );
for(i = ; i<= n ; i++)
{
v = f[i];
if(v == )//i的父节点为根节点
rootson++;//子树
else if(dfn[v] <= low[i])
Cut[v] = true;
}
for(i = ; i <= n ; i++)
if(Cut[i])
num++;
if(rootson > )
num++;
} int main()
{
int a, b;
char ch;
while(scanf("%d", &n), n)
{
Init();
while(scanf("%d", &a), a)
{
while(scanf("%d%c", &b, &ch))
{
G[a].push_back(b);
G[b].push_back(a);
if(ch == '\n')
break;
}
}
Solve();
printf("%d\n", num);
}
return ;
}

uva 315 Network(无向图求割点)的更多相关文章

  1. UVA 315 315 - Network(求割点个数)

     Network  A Telephone Line Company (TLC) is establishing a new telephone cable network. They are con ...

  2. B - Network---UVA 315(无向图求割点)

        A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connectin ...

  3. poj 1144 Network 无向图求割点

    Network Description A Telephone Line Company (TLC) is establishing a new telephone cable network. Th ...

  4. UVA 315 Network (模板题)(无向图求割点)

    <题目链接> 题目大意: 给出一个无向图,求出其中的割点数量. 解题分析: 无向图求割点模板题. 一个顶点u是割点,当且仅当满足 (1) u为树根,且u有多于一个子树. (2) u不为树根 ...

  5. 无向图求割点 UVA 315 Network

    输入数据处理正确其余的就是套强联通的模板了 #include <iostream> #include <cstdlib> #include <cstdio> #in ...

  6. UVA - 315 Network(tarjan求割点的个数)

    题目链接:https://vjudge.net/contest/67418#problem/B 题意:给一个无向连通图,求出割点的数量.首先输入一个N(多实例,0结束),下面有不超过N行的数,每行的第 ...

  7. (连通图 模板题 无向图求割点)Network --UVA--315(POJ--1144)

    链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...

  8. UVA315:Network(求割点)

    Network 题目链接:https://vjudge.net/problem/UVA-315 Description: A Telephone Line Company (TLC) is estab ...

  9. poj 1523"SPF"(无向图求割点)

    传送门 题意: 有一张联通网络,求出所有的割点: 对于割点 u ,求将 u 删去后,此图有多少个联通子网络: 对于含有割点的,按升序输出: 题解: DFS求割点入门题,不会的戳这里

随机推荐

  1. csv 文件介绍

    CSV即Comma Separate Values,这种文件格式经常用来作为不同程序之间的数据交互的格式. 具体文件格式 每条记录占一行 以逗号为分隔符 逗号前后的空格会被忽略 字段中包含有逗号,该字 ...

  2. java.lang.NoClassDefFoundError: javax/wsdl/OperationType

    You should find the javax.wsdl package inside wsdl4j.jar Check for the line starting with 'Found IBM ...

  3. sql server 数据库 ' ' 附近有语法错误

    昨天做项目时候,遇到标题的问题,代码跟踪把sql 语句 复制出来在数据库执行不了, 然后重新写个一模一样的,然后在 赋值到代码中,还是同样的错误, 就是不知道哪里出现了错误,最后 把 sql 语句写成 ...

  4. ganglia对于tomcat进程的res内存监控扩展

    ganglia是采用yum的安装,因此安装相关内容路径可能不同,但是不影响插件的扩展编写: 本次介绍的扩展是采用python脚本进行扩展,因此监控节点上需要安装python的相关插件: sudo yu ...

  5. javascript插件编写小结

    写JS插件,最好是先通过HTML方式将展示结果显示出来,然后再封装成JS插件,将其画出来.JS模板如下: (function($){ $.fn.fnName = function(options){ ...

  6. UVA 10917 Walk Through the Forest(dijkstra+DAG上的dp)

    用新模板阿姨了一天,换成原来的一遍就ac了= = 题意很重要..最关键的一句话是说:若走A->B这条边,必然是d[B]<d[A],d[]数组保存的是各点到终点的最短路. 所以先做dij,由 ...

  7. jQuery的威力

    jQuery如此之好用,和其在获取对象时使用与CSS选择器兼容的语法有很大关系,毕竟CSS选择器大家都很熟悉(关于CSS选择器可以看看十分钟搞定CSS选择器),但其强大在兼容了CSS3的选择器,甚至多 ...

  8. Ecshop ajax 局部刷新购物车功能

    1.比如我们category.dwt 里有 <a href='flow.php'><SPAN id='cart_count_all'>{insert name='cart_in ...

  9. HDU 5882 Balanced Game

    Balanced Game Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tot ...

  10. [Everyday Mathematics]20150105

    设 $f\in C^1(a,b)$ 适合 $$\bex \lim_{x\to a^+}f(x)=+\infty,\quad \lim_{x\to b^-}f(x)=-\infty, \eex$$ 并且 ...