POJ 3177(双联通分量)

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D - 双连通

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another.

Given a descri_ption of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way.

There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.

Input

Line 1: Two space-separated integers: F and R

Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.

Output

Line 1: A single integer that is the number of new paths that must be built.

Sample Input

Sample Output

Hint

Explanation of the sample:

One visualization of the paths is:

1   2   3
   +---+---+  
       |   |
       |   |
 6 +---+---+ 4
      / 5
     / 
    / 
 7 +

Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.

1   2   3
   +---+---+  
   :   |   |
   :   |   |
 6 +---+---+ 4
      / 5  :
     /     :
    /      :
 7 + - - - -

Check some of the routes:
1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2
1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4
3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7
Every pair of fields is, in fact, connected by two routes.

It's possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.

给定一个无向图，问至少加几条边使得该图双向联通，即每两个点双向可达。

自己动手画几个图会发现缩点之后(叶子节点数+1)/2就是答案。

/****************************
* author:crazy_石头
* date：2014/01/18
* algorithm:tarjan
* Pro：POJ 3177
***************************/
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>

using namespace std;

#define INF 1<<29
#define eps 1e-8
#define A system("pause")
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))

const int maxn=3000+10;
const int maxm=3000+10;

struct edge
{
	int from,to,next;
}e[maxm];
int dfn[maxn],low[maxn],belong[maxn],stack[maxn],out[maxn],head[maxn],scc,top,index,n,m,cnt;
bool instack[maxn],hash[maxn][maxn];

inline void addedge(int u,int v)
{
    e[cnt].from=u;
	e[cnt].to=v;
	e[cnt].next=head[u];
	hash[u][v]=hash[v][u]=true;
	head[u]=cnt++;
}

inline void tarjan(int u,int f)
{
	int v;
	dfn[u]=low[u]=++index;
	stack[top++]=u,instack[u]=1;
	for(int i=head[u];~i;i=e[i].next)
	{
		v=e[i].to;
		if(v==f) continue;
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
		}else if(instack[v])
        low[u]=min(low[u],dfn[v]);
	}
	if(dfn[u]==low[u])
	{
		scc++;
		do
		{
			v=stack[--top];
			instack[v]=0;
			belong[v]=scc;
		}while(u!=v);
	}
}

inline void init()
{
	int u,v;
	scc=index=top=cnt=0;
	ms(dfn,0),ms(instack,0),ms(head,-1),ms(low,0),ms(belong,0);
	rep(i,1,m)
	{
	    scanf("%d%d",&u,&v);
	    if(hash[u][v]) continue;
	    addedge(u,v);
	    addedge(v,u);
	}
	rep(i,1,n) if(!dfn[i]) tarjan(i,i);
}

inline void solve()
{
	int i,v,ans=0;;
	ms(out,0);
	rep(i,0,cnt-1)
    {
        int u=e[i].from,v=e[i].to;
        u=belong[u],v=belong[v];
		if(u!=v)    out[u]++;//记录出度；
	}
	rep(i,1,scc) if(out[i]==1) ans++;//记录叶子结点个数；
	int ret=(ans+1)/2;
	printf("%d\n",ret);
}

int main()
{
	while(~scanf("%d%d",&n,&m))
	{
		init();
		solve();//缩点后重新构图；
	}
	return 0;
}