[POJ 1144] Network【Tarjan割点】

  • 2018-01-09
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Problem:

Time Limit: 1000MS Memory Limit: 10000K

Description

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

Hint

You need to determine the end of one line.In order to make it's easy to determine,there are no extra blank before the end of each line.

Source

Central Europe 1996

Solution:

Tarjan 求割点的模板题。

割点 (Cut vertices / Articulation points) 是指在一个连通的无向图 G 的顶点集 V 中的部分顶点,其中任意一个去除后能导致 G 不再连通

利用 Tarjan 算法时间戳 dfn[] 的特性,我们可以构建出一棵解答树(可以理解成 dfs 树,在原图的基础上去掉了没有走过的边)。

这样的话,考虑到回边(即返祖边)的存在,一个节点的任意两棵子树都只能通过该节点该节点的祖先相连通。

在这棵解答树上,割点的存在分两种情况:

  1. 根节点是割点,当且仅当它至少有 2 棵子树。此时若删去该节点,它的任意两棵子树不再连通
  2. 非根节点是割点,当且仅当它的某一棵子树没有指向它的祖先的边。此时若删去该节点,该子树与其他部分不再连通

第 1 种情况很好实现。而第 2 种情况正好可以利用 Tarjan 算法的 dfn[]low[],当节点 u 到它的子节点 v 的递归返回后若

  • low[v] ≥ dfn[u]

则从 v 出发不能到达 u 的祖先,因为 u 的祖先的时间戳一定比 u 的时间戳更小

至此,Tarjan 算法求割点的任务圆满完成

伪代码参见 https://en.wikipedia.org/wiki/Biconnected_component

Code: O(T(N+M)), M为总边数 [144K, 16MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

int N;

struct Edge{
	int np;
	Edge *nxt;
};

struct Graph{
	Edge *V[102], E[10002];
	int tope;
	
	inline void clear() {tope = 0, memset(V, 0, sizeof(V));}
	
	inline void addedge(int u, int v) {E[++tope].np = v, E[tope].nxt = V[u], V[u] = &E[tope];}
} G;

int dfn[102], dfstime;
int low[102];
int vis[102], rootchild;
bool Artic[102];

inline void Tarjan(int u){
	dfn[u] = low[u] = ++dfstime, vis[u] = 1;
	for(register Edge *ne = G.V[u]; ne; ne = ne->nxt){
		if(!vis[ne->np]){
			Tarjan(ne->np);
			if(u == 1) {if(++rootchild > 1) Artic[1] = 1;}
			else{
				if(low[ne->np] >= dfn[u]) Artic[u] = 1;
				low[u] = min(low[u], low[ne->np]);
			}
		}
		else low[u] = min(low[u], dfn[ne->np]);
	}
}

int main(){
	while(scanf("%d", &N) != EOF && N){
		{
			int u, v;
			G.clear();
			while(scanf("%d", &u) != EOF && u){
				while(getchar() != '\n'){
					scanf("%d", &v);
					G.addedge(u, v), G.addedge(v, u);
				}
			}
		}  // Limit the action scope of the variables u and v
		dfstime = 0, rootchild = 0;
		memset(vis, 0, sizeof(vis));
		memset(Artic, 0, sizeof(Artic));
		for(register int i = 1; i <= N; i++)
			if(!vis[i]) Tarjan(i);
		int ans = 0;
		for(register int i = 1; i <= N; i++)
			if(Artic[i]) ans++;
		printf("%d\n", ans);
	}
	return 0;
} 

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