hdu 4612 Warm up (带有重边的无向图Tarjan+树的直径)

Warm up

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

Total Submission(s): 3947 Accepted Submission(s): 892



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

题意：有N个点，M条边，加一条边，求割边最少（有重边，所有点在原图中都是连通的）。

割边:在原图中去掉该边，原图变得不连通,这样的边成为割边。

思路：先求无向图的连通分量，缩点形成一个生成树，可知树中所有边都为割边，采用贪

心的策略可知，再加一条边，最少的割边数为 ans = 所有割边数 - 数的直径 。

因为有的图上可能有重边，这样不好处理。我们记录每条边的标号（一条无向边拆成的两

条有向边标号相同）这样就能限制不走一样的边而能走重边！

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
using namespace std;
const int maxn=200010;

struct edge
{
int u,v;
edge() {}
edge(int uu,int vv):u(uu),v(vv) {}
} a[maxn*10];

int n,m,dfs_clock,cnt,index,id[maxn],S[maxn],low[maxn],dfn[maxn],belong[maxn],num[maxn];
vector <int> G[maxn],T[maxn];
stack <int> st;
bool mark[maxn],visited[maxn*10];

void initial()
{
memset(low,0,sizeof(low));
memset(dfn,0,sizeof(dfn));
memset(num,0,sizeof(num));
memset(mark,0,sizeof(mark));
memset(belong,0,sizeof(belong));
memset(visited,0,sizeof(visited));
for(int i=0; i<maxn; i++)
{
T[i].clear();
G[i].clear();
}
while(!st.empty())  st.pop();
index=0;
dfs_clock=1;
cnt=0;
}

void input()
{
int u,v;
for(int i=0; i<m; i++)
{
scanf("%d %d",&u,&v);
G[u].push_back(cnt);
a[cnt++]=edge(u,v);
G[v].push_back(cnt);
a[cnt++]=edge(v,u);
}
}

void tarjan(int u,int fa)
{
dfn[u]=low[u]=dfs_clock++;
st.push(u);
mark[u]=1;
for(int i=0; i<G[u].size(); i++)
{
int tp=G[u][i];
if(visited[tp])  continue;
visited[tp]=visited[tp^1]=1;
int v=a[tp].v;
if(!dfn[v])
{
tarjan(v,tp);
low[u]=min(low[v],low[u]);
}
else if(mark[v])  low[u]=min(low[u],dfn[v]);
}
if(low[u]==dfn[u])
{
index++;
while(1)
{
int t=st.top();
st.pop();
belong[t]=index;
if(t==u)  break;
}
}
}

void solve1()
{
for(int i=1; i<=n; i++)
if(!dfn[i])
tarjan(i,-1);
for(int i=0; i<cnt; i+=2)
{
int p=belong[a[i].u],q=belong[a[i].v];
if(p!=q)
{
T[p].push_back(q);
T[q].push_back(p);
}
}
}

void dfs(int u,int v,int depth)
{
num[v]=depth;
for(int i=0; i<T[v].size(); i++)
{
int vv=T[v][i];
if(vv==u)  continue;
dfs(v,vv,depth+1);
}
}

void solve2()
{
int Max=-1,k=-1;
dfs(-1,1,0);
for(int i=1; i<=index; i++)  if(Max<num[i])  Max=num[i],k=i;
memset(num,0,sizeof(num));
dfs(-1,k,0);
for(int i=1; i<=index; i++)  Max=max(Max,num[i]);
printf("%d\n",index-1-Max);
}

int main()
{
while(scanf("%d %d",&n,&m)!=EOF)
{
if(n==0 && m==0)  break;
initial();
input();
solve1();
solve2();
}
return 0;
}