hdu4612 Warm up 缩点+树的直径

题意抽象后为：
给定一个无向图

问添加一条边的情况下最少能有多少个桥。

 
桥的定义：删除该边后原图变为多个连通块。 

数据规模:点数N(2<=N<=200000),边数M(1<=M<=1000000)

缩点之后求一下树的直径就好了，最优加边方案显然为连接直径的头尾。

AC代码：

#include<cstdio>
#include<cstring>
#include<queue>
#define rep(i,a,b) for(int i=a;i<=b;++i)
using namespace std;
const int MAXV=210000;
const int MAXE=2210000;
int DFN[MAXV],low[MAXV],par[MAXV],label[MAXV];
int pointer[MAXV];
int tot,cnt,m,n,Bcnt,ans;
vector<int> graph[MAXV];
int dep[MAXV];
struct Edge
{
    int to,next;
    bool vis;
    Edge() {}
    Edge(int b,int nxt,int flag) {to=b,next=nxt,vis=flag;}
}edge[MAXE];
inline void addedge(int a,int b)
{
    edge[tot]=Edge(b,pointer[a],0);
    pointer[a]=tot++;
    edge[tot]=Edge(a,pointer[b],0);
    pointer[b]=tot++;
}
void init()
{
    tot=0;
    cnt=0;Bcnt=0;
    memset(pointer,-1,sizeof(pointer));
    memset(label,0,sizeof(label));
    memset(DFN,0,sizeof(DFN));
    rep(i,1,n)
    {
        graph[i].clear();
    }
}
void tarjan(int u)
{
    DFN[u]=low[u]=++cnt;
    for(int j=pointer[u];j!=-1;j=edge[j].next)
    {
        int v=edge[j].to;
        if(edge[j].vis) continue;
        edge[j].vis=edge[j^1].vis=1;
        if(!DFN[v])
        {
            par[v]=j;
            tarjan(v);
            if(low[v]<low[u]) low[u]=low[v];
        }
        else if(low[u]>DFN[v]) low[u]=DFN[v];
    }
}
void part(int u)
{
    label[u]=Bcnt;
    for(int j=pointer[u];j!=-1;j=edge[j].next)
    {
        int v=edge[j].to;
        if(!label[v]&&edge[j].vis) part(v);
    }
}
void bfs(int s)
{
    rep(i,1,Bcnt) dep[i]=-1;
    queue<int > q;
    dep[s]=0;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();q.pop();
        for(int j=0;j<graph[u].size();++j)
        {
            int v=graph[u][j];
            if(dep[v]==-1)
            {
                dep[v]=dep[u]+1;
                q.push(v);
            }
        }
    }
}
int diameter()
{
    bfs(1);
    int ma=0,tag;
    rep(i,1,Bcnt)
    {
        if(dep[i]>ma)
        {
            tag=i;
            ma=dep[i];
        }
    }
    bfs(tag);
    ma=0;
    rep(i,1,Bcnt)
    {
        if(dep[i]>ma)
        {
            tag=i;
            ma=dep[i];
        }
    }
    return ma;
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)==2&&(n+m))
    {
        init();
        int u,v;
        rep(i,1,m)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }
        tarjan(1);
        int tmp;
        rep(i,2,n)
        {
            tmp=par[i]^1;
            u=edge[tmp].to;
            if(low[i]>DFN[u])
            {
                edge[tmp].vis=0;
                edge[tmp^1].vis=0;
            }
        }
        Bcnt=0;
        rep(i,1,n)
        {
            if(!label[i])
            {
                Bcnt++;
                part(i);
            }
        }
        rep(i,2,n)
        {
            if(!edge[par[i]].vis)
            {
                tmp=par[i]^1;
                u=edge[tmp].to;
                graph[label[u]].push_back(label[i]);
                graph[label[i]].push_back(label[u]);
            }
        }
        printf("%d
",Bcnt-1-diameter());
    }
    return 0;
}