HDU 4612 Warm up (边双联通，树的直径)

http://acm.hdu.edu.cn/showproblem.php?pid=4612

Warm up

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 3323 Accepted Submission(s): 752



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

0

Source

2013 Multi-University Training Contest 2



题意：

给出一无向图，有重边，问填一条边后使这张图中桥的数量最少是多少

分析：

找出边双联通分量（找出桥，然后去掉桥的各联通分量就是），然后缩点（遍历边），得到一颗树，找出树的直径（两次SPFA），然后桥的数量减去树的直径就行了。

注意这题有重边，所以我们要用邻接表建图。

还有HDU上2M的栈大小不够，所以我们要加上以下代码扩栈（C++提交）：

#pragma comment(linker, "/STACK:102400000,102400000")

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm 1000007<<1
#define maxn 200007
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

int pre[maxn],low[maxn],fir[maxn],e_bcc[maxn];
int u[maxm],v[maxm],nex[maxm];
bool isbridge[maxm],vise[maxm];
int n,m,e_max,dfs_clock,bcc_cnt,b_cnt;

void tarjan_dfs(int _u)
{
    pre[_u]=low[_u]=++dfs_clock;
    for (int e=fir[_u];~e;e=nex[e])
    {
        int _v=v[e];
        if (!pre[_v])
        {
            vise[e]=vise[e^1]=true;
            tarjan_dfs(_v);
            low[_u]=min(low[_u],low[_v]);
            if (pre[_u]<low[_v])    {isbridge[e]=isbridge[e^1]=true;b_cnt++;}
        }
        else
        {
            if (pre[_v]<pre[_u] && !vise[e])
            {
                vise[e]=vise[e^1]=true;
                low[_u]=min(low[_u],pre[_v]);
            }
        }
    }
}

void find_bridge()
{
    dfs_clock=b_cnt=0;
    memset(isbridge,0,sizeof isbridge);
    memset(pre,0,sizeof pre);
    memset(vise,0,sizeof vise);

    for (int i=1;i<=n;i++)
    {
        if (!pre[i])    tarjan_dfs(i);
    }
}

void dfs(int _u)
{
    e_bcc[_u]=bcc_cnt;
    for (int e=fir[_u];~e;e=nex[e])
    {
        int _v=v[e];
        if (!e_bcc[_v] && !isbridge[e])
        {
            dfs(_v);
        }
    }
}

void edge_bcc()
{
    find_bridge();//找桥
    memset(e_bcc,0,sizeof e_bcc);
    bcc_cnt=0;
    //DFS分出边双联通
    for (int i=1;i<=n;i++)
    {
        if (!e_bcc[i])
        {
            bcc_cnt++;
            dfs(i);
        }
    }

}

int d[maxn],q[maxm];
bool inq[maxn];
int f,r;

void SPFA(int _u)
{
    memset(d,0x3f,sizeof d);
    memset(inq,0,sizeof inq);
    d[_u]=0;
    f=0;r=-1;
    q[++r]=_u;
    int MAX=0;
    while (f<=r)
    {
        int x=q[f++];
        inq[x]=false;
        for (int e=fir[x];~e;e=nex[e])
        {
            if (d[v[e]]>d[u[e]]+1)
            {
                d[v[e]]=d[u[e]]+1;
                MAX=max(MAX,d[v[e]]);
                if (!inq[v[e]])
                {
                    inq[v[e]]=true;
                    q[++r]=v[e];
                }
            }
        }
    }

}

int dp[maxn];
bool visv[maxn];

int dp_dfs(int _u)
{
    if (dp[_u]!=-1) return dp[_u];

    for (int e=fir[_u];~e;e=nex[e])
    {
        if (!visv[v[e]])
        {
            visv[v[e]]=true;
            dp[_u]=max(dp[_u],dp_dfs(v[e]));
            visv[v[e]]=false;
        }

    }

    printf("dp[%d] %d\n",_u,++dp[_u]);

    return dp[_u];
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("/home/fcbruce/文档/code/t","r",stdin);
    #endif // ONLINE_JUDGE

    int x,y;

    while (scanf("%d %d",&n,&m),n||m)
    {

        memset(fir,-1,sizeof fir);
        e_max=-1;
        for (int i=0;i<m;i++)
        {
            scanf("%d %d",&x,&y);
            e_max++;
            u[e_max]=x;v[e_max]=y;
            nex[e_max]=fir[x];
            fir[x]=e_max;
            e_max++;
            u[e_max]=y;v[e_max]=x;
            nex[e_max]=fir[y];
            fir[y]=e_max;
        }

        edge_bcc();

        int bccemax=-1;
        memset(fir,-1,sizeof fir);
        for (itn e=0;e<=e_max;e++)
        {
            if (e_bcc[u[e]]==e_bcc[v[e]])   continue;
            bccemax++;
            u[bccemax]=e_bcc[u[e]];v[bccemax]=e_bcc[v[e]];
            nex[bccemax]=fir[u[bccemax]];
            fir[u[bccemax]]=bccemax;
        }

        SPFA(1);
        int MAXV=-1;
        for (int i=1;i<=bcc_cnt;i++)
            if (MAXV==-1 || d[MAXV]<d[i])   MAXV=i;

        SPFA(MAXV);
        MAXV=-1;
        for (int i=1;i<=bcc_cnt;i++)
            if (MAXV==-1 || d[MAXV]<d[i])   MAXV=i;

        printf("%d\n",b_cnt-d[MAXV]);

    }

    return 0;
}