Redundant Paths-POJ3177（并查集+双连通分量）

Redundant Paths

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 description 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

7 7
1 2
2 3
3 4
2 5
4 5
5 6
5 7

Sample Output

2

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.

Source

USACO 2006 January Gold

题意：

给定一个无向连通图，判断至少加多少的边，才能使任意两点之间至少有两条的独立的路（没有公共的边，但可以经过同一个中间的顶点）。

思路：

在同一个双连通分量里的所有的点可以看做一个点，收缩后，新图是一棵树，树的边边是原图的桥。现在问题转化为“在树中至少添加多少条边能使图变成边双连通图”，即添加的边的个数=（树中度为一的节点数目+1）/2；

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define LL long long 

using namespace std;

const int MaxN = 10100;

const int MaxM = 5100;
// 建图
typedef struct node
{
	int v;
	int next;
}Line;

Line Li[MaxN*2]; 

int Belong[MaxM];//判断所属的双连通的集合
//回到最优先遍历度最小的节点和记录点DFS的顺序
int low[MaxM],dfn[MaxM];

int Head[MaxM],top;

int pre[MaxM]; //并查集

int vis[MaxM],Num;//标记点的状态

int Bridge[MaxM][2],NumB;//记录图中的桥

void AddEdge(int u,int v)// 建边
{
	Li[top].v=v; Li[top].next=Head[u];
	Head[u]=top++;

	Li[top].v=u; Li[top].next=Head[v];
	Head[v]=top++;
}

int Find(int x)// 路径压缩
{
	return pre[x]==-1?x:pre[x]=Find(pre[x]);
}

void Union(int x,int y)
{
	int Fx = Find(x);

	int Fy = Find(y);

	if(Fx!=Fy)
	{
		pre[Fx] = Fy;
	}
}

void dfs(int u,int father)
{
	low[u]=dfn[u]=Num++;

	vis[u]=1;//表示已经遍历但是没有遍历完子女

	for(int i=Head[u];i!=-1;i=Li[i].next)
	{
		int j=Li[i].v;

		if(vis[j]==1&&j!=father)// 回到的祖先
		{
			low[u]=min(low[u],dfn[j]);
		}
		if(vis[j]==0)
		{
			dfs(j,u);

			low[u]=min(low[j],low[u]);

			if(low[j]<=dfn[u])//说明是双连分量中的边
			{
				Union(u,j);
			}
			if(low[j]>dfn[u])//桥
			{
				Bridge[NumB][0]=j;
				Bridge[NumB++][1]=u;
			}
		}
	}
	vis[u]=2;
}

int main()
{
	int n,m;

	while(~scanf("%d %d",&n,&m))
	{

		top=0;

		int u,v;

		memset(Head,-1,sizeof(Head));

		memset(pre,-1,sizeof(pre));
		for(int i=1;i<=m;i++)
		{
			scanf("%d %d",&u,&v);

			AddEdge(u,v);
		}

		memset(vis,0,sizeof(vis));

		memset(Belong ,-1 ,sizeof(Belong));

		Num = 0 ; NumB=0 ;

		dfs(1,0);

		Num = 0;
		for(int i=1;i<=n;i++)
		{
			int k=Find(i);//给双连通分量集合编号

			if(Belong[k]==-1)
			{
				Belong[k]=++Num;
			}
			Belong[i]=Belong[k];
		}
		memset(vis,0,sizeof(vis));

		for(int k=0;k<NumB;k++)
		{
			int i=Bridge[k][0];

			int j=Bridge[k][1];

			vis[Belong[i]]++;

			vis[Belong[j]]++;
		}

		int ans = 0;

		for(int i=1;i<=n;i++)
		{
			if(vis[i]==1)
			{
				ans++;
			}
		}
		printf("%d\n",(ans+1)/2);
	}
	return 0;
}