hdoj 4612 Warm up【双连通分量求桥&&缩点建新图求树的直径】

Warm up

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 5093    Accepted Submission(s): 1131

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：n个点m条边，问新建一条边，可以让桥的数量达到最少，输出最少的桥数

 

题解：先求出所有的桥的数量，将桥连接的各个图进行缩点，根据树的直径的求法，求出最长的路径，用桥数量减去树的直径

//scc代表 双联通分量  写习惯了 顺手就写成scc了也懒得改了 
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#define MAXM 2000100
#define MAX 200100
#define INF 0x7fffff
using namespace std;
int n,m,bridge,sum;
int low[MAX],dfn[MAX];
int head[MAX],ans,age;
int sccno[MAX];//代表当前点属于哪个双连通分量 
int dfsclock,scccnt;
vector<int>newmap[MAX];//储存缩点后的新图 
int instack[MAX];//标记当前点是否入栈 
int dis[MAX];//求树的直径时记录路径的长度 
int vis[MAX];//求树的直径时标记是否入队列 
stack<int>s;
struct node
{
	int beg,end,next;
}edge[MAXM];
void init()
{
	ans=0;
	bridge=0;
	memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
    edge[ans].beg=u;
    edge[ans].end=v;
    edge[ans].next=head[u];
	head[u]=ans++;
}
void getmap()
{
	int a,b;
	while(m--)
	{
		scanf("%d%d",&a,&b);
		add(a,b);
		add(b,a);
	}
}
void tarjan(int u,int fa)
{
	int v,i;
	low[u]=dfn[u]=++dfsclock;
	instack[u]=1;
	s.push(u);
	int flag=1;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].end;
		if(flag&&v==fa)//判重边 
		{
			flag=0;
			continue;
		}		    
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
			if(dfn[u]<low[v])
				bridge++;
		}
		else if(instack[v])
		    low[u]=min(low[u],dfn[v]);
	}
    if(dfn[u]==low[u])
    {
    	scccnt++;
    	while(1)
    	{
    		v=s.top();
    		s.pop();
			instack[v]=0;
			sccno[v]=scccnt;
			if(v==u)
			break; 
    	}
    }
}
void find()
{
    int i;
	memset(low,0,sizeof(low));
	memset(sccno,0,sizeof(sccno));
	memset(dfn,0,sizeof(dfn));
	memset(instack,0,sizeof(instack));
	dfsclock=scccnt=0;
	for(i=1;i<=n;i++)
	{
		if(!dfn[i])
			tarjan(i,-1);
	}	
}
void suodian()
{
	int u,v,i,j;
	for(i=1;i<=scccnt;i++)
		newmap[i].clear();
	for(i=0;i<ans;i=i+2)
	{
		u=sccno[edge[i].beg];
		v=sccno[edge[i].end];
		if(u!=v)
		{
			newmap[u].push_back(v);
			newmap[v].push_back(u);
		}
	}
}
void bfs(int beg)
{
	queue<int>q;
	memset(dis,0,sizeof(dis));
	memset(vis,0,sizeof(vis));
	int i,j;
	while(!q.empty())
	    q.pop();
	sum=0;
	age=beg;
	vis[beg]=1;
	q.push(beg);
	int u;
	while(!q.empty())
	{
		u=q.front();
		q.pop();
		for(i=0;i<newmap[u].size();i++)
		{
			if(!vis[newmap[u][i]])
			{
				dis[newmap[u][i]]=dis[u]+1;
				vis[newmap[u][i]]=1;
				q.push(newmap[u][i]);
				if(sum<dis[newmap[u][i]])
				{
					sum=dis[newmap[u][i]];
					age=newmap[u][i];
				}
			}
		}
	}
}
void solve()
{
	int i,j;
	bfs(1);
	bfs(age);
	printf("%d
",bridge-sum);
	//printf("%d
%d
",bridge,sum); 
}
int main()
{
	while(scanf("%d%d",&n,&m),n|m)
	{
		init();
		getmap();
		find();
		//printf("%d#
",bridge);
		suodian();
		solve();
	}
	return 0;
}