HDU 4612 Warm up（边双联通求树的直径）

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 41 21 31 42 30 0 

 

Sample Output

0

 

Source

2013 Multi-University Training Contest 2

 

题意：加入一条边后的最小桥数。

边双联通缩点求树的直径，将直径两端连接就是减少的最多桥数。

重边问题：注意重边。又有就是要扩栈！！

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int maxn=200000+100;const int maxm=3000000+100;struct node{    int to,next;    bool col;//为桥}e[maxm],e1[maxm];int head[maxn],cnt,cntE,cnte;int DFN[maxn],low[maxn];int s[maxn],instack[maxn];int idex,top,bridge;int belong[maxn],h[maxn];int d[maxn],vis[maxn];//为割;删除点后增加的联通快int n,m;void init(){    cnt=top=idex=0;    cnte=bridge=cntE=0;    CLEAR(head,-1);    CLEAR(DFN,0);    CLEAR(low,0);    CLEAR(instack,0);    CLEAR(belong,0);    CLEAR(h,-1);    CLEAR(vis,0);    CLEAR(d,0);}void addedge(int u,int v){    e[cntE].to=v;e[cntE].next=head[u];    e[cntE].col=false;head[u]=cntE++;}void addedge1(int u,int v){    e1[cnte].to=v;e1[cnte].next=h[u];    h[u]=cnte++;}void Tarjan(int u,int pre){    int v;    low[u]=DFN[u]=++idex;    s[top++]=u;    instack[u]=1;    int pre_num=0;    for(int i=head[u];i!=-1;i=e[i].next)    {        v=e[i].to;        if(v==pre&&!pre_num)        {            pre_num++;            continue;        }        if(!DFN[v])        {            Tarjan(v,u);            if(low[u]>low[v]) low[u]=low[v];            if(low[v]>DFN[u])//桥            {                bridge++;                e[i].col=true;                e[i^1].col=true;            }        }        else if(instack[v]&&low[u]>DFN[v])            low[u]=DFN[v];    }    if(low[u]==DFN[u])    {        cnt++;        do{            v=s[--top];            instack[v]=0;            belong[v]=cnt;        }while(v!=u);    }}void dfs(int u,int depth){    d[u]=depth;    vis[u]=1;    for(int i=h[u];i!=-1;i=e1[i].next)    {        int v=e1[i].to;        if(!vis[v])            dfs(v,depth+1);    }}void work(){    REPF(i,1,n)       if(!DFN[i])  Tarjan(i,-1);    REPF(u,1,n)        for(int i=head[u];i!=-1;i=e[i].next)        {            int v=e[i].to;            if(belong[u]!=belong[v])            {                addedge1(belong[u],belong[v]);                addedge1(belong[v],belong[u]);            }        }    dfs(1,0);    int pos=1;    for(int i=1;i<=cnt;i++)        if(d[i]>d[pos])  pos=i;    CLEAR(vis,0);    CLEAR(d,0);    dfs(pos,0);    int ans=0;    REPF(i,1,cnt)  ans=max(ans,d[i]);    printf("%d\n",bridge-ans);}int main(){    int u,v;    while(~scanf("%d%d",&n,&m)&&(n+m))    {        init();        REPF(i,1,m)        {            scanf("%d%d",&u,&v);            if(u==v)  continue;            addedge(u,v);            addedge(v,u);        }        work();    }    return 0;}