Warm up

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 5184 Accepted Submission(s): 1159

Problem Description

　　N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels.
　　If we can isolate some planets from others by breaking only one channel , the channel is called a bridge of the transportation system.
People don't like to be isolated. So they ask what's the minimal number of bridges they can have if they decide to build a new channel.
　　Note that there could be more than one channel between two planets.

Input

　　The input contains multiple cases.
　　Each case starts with two positive integers N and M , indicating the number of planets and the number of channels.
　　(2<=N<=200000, 1<=M<=1000000)
　　Next M lines each contains two positive integers A and B, indicating a channel between planet A and B in the system. Planets are numbered by 1..N.
　　A line with two integers '0' terminates the input.

Output

　　For each case, output the minimal number of bridges after building a new channel in a line.

Sample Input

4 4
1 2
1 3
1 4
2 3
0 0

Sample Output

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<stack>

#include<queue>

#include<cstdlib>

using namespace std;

const int N=;

struct node

{

    int next,to;

}edge[N*],Edge[N*];

int Stack[N],low[N],dfn[N],belong[N],sum,Time,top;

int Head[N],head[N],ans,Ans,Max,dis[N],vis[N],End;

void Inn()

{

    memset(Stack,,sizeof(Stack));

    memset(low,,sizeof(low));

    memset(dfn,,sizeof(dfn));

    memset(Head,-,sizeof(Head));

    memset(head,-,sizeof(head));

    memset(belong,,sizeof(belong));

    memset(dis,,sizeof(dis));

    sum=Time=top=ans=Ans=Max=End=;

}

void add(int from,int to)

{

    edge[ans].to=to;

    edge[ans].next=head[from];

    head[from]=ans++;

}

void Add(int from,int to)

{

    Edge[Ans].to=to;

    Edge[Ans].next=Head[from];

    Head[from]=Ans++;

}

void Tarjin(int u,int f)

{

    int k=,v;

    low[u]=dfn[u]=++Time;

    Stack[top++]=u;

    for(int i=head[u];i!=-;i=edge[i].next)

    {

        v=edge[i].to;

        if(v==f && !k)

        {

            k++;

            continue;

        }

        if(!dfn[v])

        {

            Tarjin(v,u);

            low[u]=min(low[u],low[v]);

        }

        else

            low[u]=min(low[u],dfn[v]);

    }

    if(dfn[u]==low[u])

    {

        sum++;

        do

        {

            v=Stack[--top];

            belong[v]=sum;

        }while(v!=u);

    }

}

void dfs(int s)

{

    queue<int>Q;

    int u,v;

    memset(vis,,sizeof(vis));

    dis[s]=;

    vis[s]=;

    Q.push(s);

    while(Q.size())

    {

        u=Q.front();

        Q.pop();

        for(int i=Head[u];i!=-;i=Edge[i].next)

        {

            v=Edge[i].to;

            if(!vis[v])

            {

                vis[v]=;

                dis[v]=dis[u]+;

                Q.push(v);

                if(Max<dis[v])

                {

                    Max=dis[v];

                    End=v;

                }

            }

        }

    }

}

void slove(int n)

{

    int SUM=;

    Tarjin(,);

    for(int i=;i<=n;i++)

    {

        for(int j=head[i];j!=-;j=edge[j].next)

        {

            int u=belong[i];

            int v=belong[edge[j].to];

            if(u!=v)

            {

                Add(u,v);

                SUM++;

            }

        }

    }

    SUM/=;

    dfs();

    dfs(End);

    printf("%d\n",SUM-Max);

}

int main()

{

    int n,m,a,b;

    while(scanf("%d %d",&n,&m),n+m)

    {

        Inn();

        while(m--)

        {

            scanf("%d %d",&a,&b);

            add(a,b);

            add(b,a);

        }

        slove(n);

    }

    return ;

}

巴特西

Warm up-HUD4612(树的直径+Tarjin缩点)

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