POJ 1144 Network（tarjan 求割点个数）

Network

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17016		Accepted: 7635

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

 

分析：

好惭愧啊，现在才会求割点，学习了一个新的算法，tarjan算法，但现在只会用他来求割点和割边....

tarjan学习是参考的这位大佬的博客，写得真的简单易懂

https://www.cnblogs.com/nullzx/archive/2017/12/04/7968110.html

 

code：

邻接矩阵实现：

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};

int getval()
{
    int ret(0);
    char c;
    while((c=getchar())==' '||c=='
'||c=='
');
    ret=c-'0';
    while((c=getchar())!=' '&&c!='
'&&c!='
')
        ret=ret*10+c-'0';
    return ret;
}

#define max_v 1005
int dfn[max_v];
int low[max_v];
int vis[max_v];
int G[max_v][max_v];

int depth,n,m,root,rt_cnt;

void init()
{
    root=1;
    depth=1;
    rt_cnt=0;
    memset(dfn,-1,sizeof(dfn));
    memset(G,0,sizeof(G));
    memset(vis,0,sizeof(vis));
    memset(low,0,sizeof(low));
}

void tarjan(int cur,int pa)
{
    dfn[cur]=low[cur]=depth++;
    for(int i=1;i<=n;i++)
    {
        if(G[cur][i])
        {
            if(dfn[i]==-1)//i没有访问过
            {
                tarjan(i,cur);
                low[cur]=min(low[cur],low[i]);//逐步回溯更新访问过的父节点的low

                if(cur==root)
                    rt_cnt++;//统计和根结点直接相连的点的个数，来确定根结点是不是割点
                else if(low[i]>=dfn[cur])
                    vis[cur]=1;//标记当前cur点为割点

            }else if(i!=pa)//访问过，但不是父节点，更新low
            {
                low[cur]=min(low[cur],dfn[i]);
            }
        }
    }
}
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        init();
        int temp;
        while(~scanf("%d",&temp)&&temp)
        {
            while(getchar()!='
')
            {
                int t;
                scanf("%d",&t);
                G[temp][t]=G[t][temp]=1;
            }
        }
        tarjan(1,root);
        int cnt=0;
        for(int i=2;i<=n;i++)
            if(vis[i])
            cnt++;
        if(rt_cnt>1)
            cnt++;
        printf("%d
",cnt);
    }
    return 0;
}

邻接链表实现：

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};

int getval()
{
    int ret(0);
    char c;
    while((c=getchar())==' '||c=='
'||c=='
');
    ret=c-'0';
    while((c=getchar())!=' '&&c!='
'&&c!='
')
        ret=ret*10+c-'0';
    return ret;
}

#define max_v 105
int dfn[max_v];
int low[max_v];
int vis[max_v];
vector<int> vv[max_v];
int depth,n,m,root,rt_cnt;

void init()
{
    root=1;
    depth=1;
    rt_cnt=0;
    memset(dfn,-1,sizeof(dfn));
    memset(vis,0,sizeof(vis));
    memset(low,0,sizeof(low));
    //for(int i=1;i<=n;i++)
       // vv[i].clear();
    memset(vv,0,sizeof(vv));
}

void tarjan(int cur,int pa)
{
    dfn[cur]=low[cur]=depth++;
    for(int i=0;i<vv[cur].size();i++)
    {
        int temp=vv[cur][i];
        if(dfn[temp]==-1)
        {
            tarjan(temp,cur);
            low[cur]=min(low[cur],low[temp]);

            if(cur==root)
                rt_cnt++;
            else if(low[temp]>=dfn[cur])
                vis[cur]=1;

        }else if(temp!=pa)
        {
            low[cur]=min(low[cur],dfn[temp]);
        }
    }
}

int main()
{
    while(~scanf("%d",&n)&&n)
    {
        init();
        int temp;
        while(~scanf("%d",&temp)&&temp)
        {
            while(getchar()!='
')
            {
                int t;
                scanf("%d",&t);
                vv[temp].push_back(t);
                vv[t].push_back(temp);
            }
        }
        tarjan(1,root);
        int cnt=0;
        for(int i=2;i<=n;i++)
            if(vis[i])
            cnt++;
        if(rt_cnt>1)
            cnt++;
        printf("%d
",cnt);
    }
    return 0;
}