poj1144-Network （求割点）

最新推荐文章于 2024-10-27 18:03:27 发布

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最新推荐文章于 2024-10-27 18:03:27 发布

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分类专栏： Tarjan

本文链接：https://blog.youkuaiyun.com/weixin_41380961/article/details/83900631

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本文介绍了一个电话线路公司面对网络连通性问题的解决方法，通过使用割点算法来找出网络中关键的地方，这些地方的故障可能导致整个网络的连通性问题。文章详细解释了如何使用Tarjan算法来识别网络中的割点，并提供了完整的代码实现。

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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. Fromtimetotimethepowersupplyfailsataplaceandthentheexchangedoesnotoperate. The officials from TLC realized that in such a case it can happen that besides the fact that the place withthefailureisunreachable, thiscanalsocausethatsomeotherplacescannotconnecttoeachother. 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 numberofaplacefollowedbythenumbersofsomeplacestowhichthereisadirectlinefromthisplace. 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 1 2 3 4

2 1 3

5 4 6 2

0 Sample Output

割点讲解博客：https://blog.youkuaiyun.com/zsyz_ZZY/article/details/79907335

https://www.cnblogs.com/en-heng/p/4002658.html

#include<algorithm>
#include<string.h>
#include<stdio.h>
#define M 110
using namespace std;
int mp[M][M],n,x,y,LOW[M],DFN[M],indox,ans,ap[M],pan;
void init()
{
    memset(DFN,-1,sizeof(DFN));
    memset(LOW,-1,sizeof(LOW));
    memset(mp,0,sizeof(mp));
    memset(ap,0,sizeof(ap));
    indox=ans=pan=0;
}
void tarjan(int u)
{
    DFN[u]=LOW[u]=++indox;
    for(int i=1; i<=n; i++)
    {
        if(mp[u][i])
        {
            if(DFN[i]==-1)
            {
                tarjan(i);
                LOW[u]=min(LOW[u],LOW[i]);
                if((LOW[i]>=DFN[u])&&u!=1&&!ap[u])
                {
                    ap[u]=1;
                    ans++;
                }
                else if(u==1)
                {
                    pan++;
                }
            }
            else
            {
                LOW[u]=min(LOW[u],DFN[i]);
            }
        }
    }
}
int main()
{
    while(~scanf("%d",&n),n)
    {
        init();
        while(~scanf("%d",&x),x)
        {
            while(getchar()!='\n')
            {
                scanf("%d",&y);
                mp[x][y]=1;
                mp[y][x]=1;
            }
        }
        tarjan(1);
        if(pan>1) ans++;
        printf("%d\n",ans);
    }
}