poj1144 tarjan求割点 裸题

Network

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11684		Accepted: 5422

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

---

//tarjan算法求割点模板
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>

const int inf = 0x3f3f3f;
const int MAXN = 1e2+10;

using namespace std;

vector<int> adj[MAXN];
int low[MAXN]; //能访问最先祖先结点位置
int dfn[MAXN]; //记录节点位置
int cur[MAXN]; //记录是否为割点
int rt,rt_sons; //根节点、根节点的子树数
int n;

void init(){
    rt_sons = 0;
    rt = 1;
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(cur,0,sizeof(cur));
    for(int i=1;i<=n;i++)
        adj[i].clear();
}

void dfs(int u,int degree){
    dfn[u] = low[u] = degree;
    int v;
    for(int i=0;i<adj[u].size();i++){
        v = adj[u][i];
        if(!dfn[v]){ //后继结点没有被访问过，说明没有形成一个环
            dfs(v,degree+1);
            if(u==rt)rt_sons++;
            else{
                if(low[v]>=dfn[u]) //后继结点搜不到比该结点更早的结点，则说明该节点是割点
                    cur[u] = 1;
                low[u] = min(low[u],low[v]); //注意时刻更新结点的能搜索父节点的位置，后结点可能会形成一个环，搜索到比该结点更早的位置
            }
        }
        else{ //后继结点已经访问过了，也就是回环的情况，能访问的位置必然更新
            low[u] = min(low[u],dfn[v]); 
        }

    }
}

int main()
{
    int u,v;
    while(scanf("%d",&n),n){
        init();
        while(scanf("%d",&u),u){
            while(getchar()!='
'){
                scanf("%d",&v);
                adj[u].push_back(v);
                adj[v].push_back(u);
            }
        }
        dfs(rt,1);
        int sum=0;
        if(rt_sons>=2)cur[1] = 1; //y有两个子树的根是割点
        for(int i=1;i<=n;i++){
            if(cur[i])sum++;
        }
        /*cout<<"debug"<<endl;
        for(int i=1;i<=n;i++){
            cout<<cur[i]<<" ";
        }
        cout<<endl;
        cout<<"debug"<<endl;*/
        cout<<sum<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}