UVA315 Network —— 割点
题目链接:https://vjudge.net/problem/UVA-315
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 most N 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
题解:
模板题,求割点的个数。
注意:由于根节点没有父亲结点,所以在求割点的时候需要分开处理。
代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e2+; struct Edge
{
int to, next;
}edge[MAXN*MAXN*];
int tot, head[MAXN]; int Index, DFN[MAXN], Low[MAXN];
bool cut[MAXN]; void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
} void Tarjan(int u, int pre)
{
DFN[u] = Low[u] = ++Index;
int son = ;
for(int i = head[u]; i!=-; i = edge[i].next)
{
int v = edge[i].to;
if(v==pre) continue;
if(!DFN[v])
{
son++;
Tarjan(v, u);
Low[u] = min(Low[u], Low[v]);
if(u!=pre && Low[v]>=DFN[u])
cut[u] = true;
}
else
Low[u] = min(Low[u], DFN[v]);
} if(u==pre && son>) cut[u] = true;
} void init()
{
tot = ;
memset(head, -, sizeof(head)); Index = ;
memset(DFN, , sizeof(DFN));
memset(Low, , sizeof(Low));
memset(cut, false, sizeof(cut));
} int main()
{
int n;
while(scanf("%d", &n) &&n)
{
init();
int u, v;
while(scanf("%d", &u) && u)
{
while(getchar()!='\n')
{
scanf("%d", &v);
addedge(u, v);
addedge(v, u);
}
} Tarjan(, );
int ans = ;
for(int i = ; i<=n; i++)
if(cut[i]) ans++; printf("%d\n", ans);
}
}
UVA315 Network —— 割点的更多相关文章
- uva-315.network(连通图的割点)
本题大意:求一个无向图额割点的个数. 本题思路:建图之后打一遍模板. /**************************************************************** ...
- [UVA315]Network(tarjan, 求割点)
题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem ...
- UVA315 Network 连通图割点
题目大意:有向图求割点 题目思路: 一个点u为割点时当且仅当满足两个两个条件之一: 1.该点为根节点且至少有两个子节点 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 ...
- poj 1144 Network(割点)
题目链接: http://poj.org/problem?id=1144 思路分析:该问题要求求出无向联通图中的割点数目,使用Tarjan算法即可求出无向联通图中的所有的割点,算法复杂度为O(|V| ...
- poj 1144 Network(割点 入门)
Network Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 10907 Accepted: 5042 Descript ...
- UVA315 Network
割点的概念:对于无向图,删除这个点与其相连的边,整个图的连通分量个数增加. 对于无向图的tarjan算法,必须要设前驱~ 求割点的模板~ #include<cstdio> #include ...
- 连通图(Tarjan算法) 专题总结
一.题目类型: 1.有向图的强连通分量: POJ1236 Network of Schools HDU1269 迷宫城堡 2.割点 & 割边: UESTC - 900 方老师炸弹 UVA315 ...
- UVA315:Network(求割点)
Network 题目链接:https://vjudge.net/problem/UVA-315 Description: A Telephone Line Company (TLC) is estab ...
- Network -UVa315(连通图求割点)
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=5&page=sh ...
随机推荐
- 洛谷 P3387 【模板】缩点 DAGdp学习记
我们以洛谷P3387 [模板]缩点 来学习DAGdp 1.这道题的流程 //伪代码 for i->n if(i未被遍历) tarjan(i) 缩点() DAGdp() 完成 首先tarjan这部 ...
- Computers(线性DP)
描述 Everybody is fond of computers, but buying a new one is always a money challenge. Fortunately, th ...
- NYOJ301-递推求值
递推求值 nyoj上矩阵专题里的10道题水了AC率最高的5道,惭愧,还不是完全自己写的,用了几乎两周的时间.模板题我是有自信写出来的,但对于高级一点的矩阵构造,我还是菜的抠脚. 这题感谢MQL大哥和她 ...
- 什么是Service Mesh?
转至大佬宋净明的博客:https://jimmysong.io/posts/what-is-a-service-mesh/ Service mesh 又译作 “服务网格”,作为服务间通信的基础设施层. ...
- POJ 1236 学校网络间的强连通
题目大意: N个学校之间有单向的网络,每个学校得到一套软件后,可以通过单向网络向周边的学校传输.问题1:初始至少需要向多少个学校发放软件,使得网络内所有的学校最终都能得到软件.问题2:至少需要添加几条 ...
- PHP复制和移动目录
<?php //重命名一个文件或目录 rename("phpmyadmin", "phpadmin");//重命名成phpadmin /* * $dirs ...
- Linux内核设计与实现——读书笔记2:进程管理
1.进程: (1)处于执行期的程序,但不止是代码,还包括各种程序运行时所需的资源,实际上进程是正在执行的 程序的实时结果. (2)程序的本身并不是进程,进程是处于执行期的程序及其相关资源的总称. (3 ...
- 并发编程——IO模型
前言 同步(synchronous):一个进程在执行某个任务时,另外一个进程必须等待其执行完毕,才能继续执行 #所谓同步,就是在发出一个功能调用时,在没有得到结果之前,该调用就不会返回.按照这个定义, ...
- 进程&进程池
进程 服务器中, s.listen(n) n不能无限大,以为内存不可能无限大,n表示内存同一时间接纳的等待连接数,可以看成一个(队列),取出一个拿去建立连接,然后再放进一个,队列中一直保持n个连接 请 ...
- P1195 口袋的天空 洛谷
https://www.luogu.org/problem/show?pid=1195 题目背景 小杉坐在教室里,透过口袋一样的窗户看口袋一样的天空. 有很多云飘在那里,看起来很漂亮,小杉想摘下那样美 ...