(边双联通+树直径) hdu 4612

Warm up

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4437 Accepted Submission(s): 1001

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

Author

SYSU

题意

一个无向图，现在要添加一条边，使得桥的数量最少并输出。

首先添加的边肯定是树的直径两个端点，所以我们只需要求出总的桥-树直径就可以了啊

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<stack>
using namespace std;
vector<int> e[200005],mp[200005];
int n,m,use[200005],top,newflag,Dfs[200005],low[200005],isstack[200005];
int dp[200005],vis[200005],ans,bridge;
stack<int> s;
void init()
{
    ans=0,bridge=0;
    memset(use,0,sizeof(use));
    memset(Dfs,0,sizeof(Dfs));
    memset(low,0,sizeof(low));
    memset(isstack,0,sizeof(isstack));
    top=newflag=0;
    while(!s.empty())
        s.pop();
    for(int i=1;i<=n;i++)
        e[i].clear(),mp[i].clear();
}
int bfs(int x)
{
    int pos;
    pos=x;
    memset(vis,0,sizeof(vis));
    memset(dp,0,sizeof(dp));
    queue<int> q;
    q.push(x);
    vis[x]=1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=0;i<mp[u].size();i++)
        {
            int v=mp[u][i];
            if(vis[v]) continue;
            vis[v]=1;
            dp[v]=dp[u]+1;
            q.push(v);
            if(dp[v]>ans)
            {
                ans=dp[v];
                pos=v;
            }
        }
    }
    return pos;
}
void tarjan(int u,int father)
{
    Dfs[u]=low[u]=++top;
    s.push(u);
    isstack[u]=1;
    int cnt=0;
    for(int i=0;i<e[u].size();i++)
    {
        int v=e[u][i];
        if(!Dfs[v])
        {
            tarjan(v,u);
            low[u]=min(low[u],low[v]);
            if(low[v]>Dfs[u])
                bridge++;
        }
        else if(v==father)
        {
            if(cnt) low[u]=min(low[u],Dfs[v]);
            cnt++;
        }
        else if(isstack[v])
            low[u]=min(low[u],Dfs[v]);
    }
    if(low[u]==Dfs[u])
    {
        newflag++;
        int x;
        do
        {
            x=s.top();
            s.pop();
            isstack[x]=0;
            use[x]=newflag;
        }while(x!=u);
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)
            break;
        init();
        for(int i=1;i<=m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            e[x].push_back(y);
            e[y].push_back(x);
        }
        for(int i=1;i<=n;i++)
        {
            if(!Dfs[i])
                tarjan(i,-1);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<e[i].size();j++)
            {
                int u,v;
                u=i,v=e[i][j];
                if(use[u]!=use[v])
                    mp[use[u]].push_back(use[v]);
            }
        }
        bfs(bfs(1));
        printf("%d
",bridge-ans);
    }
    return 0;
}