poj 3352 Road Construction【边双连通求最少加多少条边使图双连通&&缩点】

Road Construction

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10141		Accepted: 5031

Description

It's almost summer time, and that means that it's almost summer construction time! This year, the good people who are in charge of the roads on the tropical island paradise of Remote Island would like to repair and upgrade the various roads that lead between the various tourist attractions on the island.

The roads themselves are also rather interesting. Due to the strange customs of the island, the roads are arranged so that they never meet at intersections, but rather pass over or under each other using bridges and tunnels. In this way, each road runs between two specific tourist attractions, so that the tourists do not become irreparably lost.

Unfortunately, given the nature of the repairs and upgrades needed on each road, when the construction company works on a particular road, it is unusable in either direction. This could cause a problem if it becomes impossible to travel between two tourist attractions, even if the construction company works on only one road at any particular time.

So, the Road Department of Remote Island has decided to call upon your consulting services to help remedy this problem. It has been decided that new roads will have to be built between the various attractions in such a way that in the final configuration, if any one road is undergoing construction, it would still be possible to travel between any two tourist attractions using the remaining roads. Your task is to find the minimum number of new roads necessary.

Input

The first line of input will consist of positive integers n and r, separated by a space, where 3 ≤ n ≤ 1000 is the number of tourist attractions on the island, and 2 ≤ r ≤ 1000 is the number of roads. The tourist attractions are conveniently labelled from 1 ton. Each of the following r lines will consist of two integers, v and w, separated by a space, indicating that a road exists between the attractions labelled v and w. Note that you may travel in either direction down each road, and any pair of tourist attractions will have at most one road directly between them. Also, you are assured that in the current configuration, it is possible to travel between any two tourist attractions.

Output

One line, consisting of an integer, which gives the minimum number of roads that we need to add.

Sample Input

Sample Input 1
10 12
1 2
1 3
1 4
2 5
2 6
5 6
3 7
3 8
7 8
4 9
4 10
9 10

Sample Input 2
3 3
1 2
2 3
1 3

Sample Output

Output for Sample Input 1
2

Output for Sample Input 2
0

题意：有n个旅游景点和r条路，任意两个景点之间都间接或直接的连在一起，但是有时候道路施工时我们无法走这条路，也就是说两个景点之间的道路断了，问现在最少添加多少条通道，使任意两个景点之间都不止一条通道

题解：显而易见此题是让求最少添加多少条路使图双连通，因为所有的点都是相连的，只是不是双连通，我们可以把原始数据看做是一个双连通图去掉了几条边所变成的图形，现在我们需要将这几条边加上去，一个双连通图其度数至少为2，所以当我们遇见度数为1的点时记录下来，最后度数为1的点的总数（sum+1）/2就是最后结果

注意：输入输出真坑，本来以为每组输入和输出前都要输出字符串的  结果一直wa

#include<stdio.h>
#include<string.h>
#include<stack>
#include<algorithm>
#define MAX 21000
#include<vector>
#define MAXM 2001000
#define INF 0x7ffffff
using namespace std;
int n,m,num,bridge;
int head[MAX],ans;
int in[MAX];
int low[MAX],dfn[MAX];
int instack[MAX],iscut[MAX];
int addbcc[MAX];
int dfsclock,bccno[MAX];
int bcccnt;
stack<int>s;
vector<int>newmap[MAX];
vector<int>bcc[MAX];
struct node
{
	int beg,end,next;
}edge[MAXM];
void init()
{
	ans=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,i;
	while(m--)
	{
		scanf("%d%d",&a,&b);
		add(a,b);
		add(b,a);
	}
}
void tarjan(int u,int fa)
{
	int i,j,v;
	s.push(u);
	instack[u]=1;
	low[u]=dfn[u]=++dfsclock;
	int son=0;
	int flag=1;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].end;
		if(v==fa&&flag)//去重边
		{
			flag=0;
			continue;
		}
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
			if(low[v]>dfn[u])//是桥
			    bridge++;
		}
		else
		    low[u]=min(low[u],dfn[v]);
	}
    if(low[u]==dfn[u])
    {
    	bcccnt++;
    	while(1)
    	{
    		v=s.top();
    		s.pop();
    		instack[v]=0;
    		bccno[v]=bcccnt;
    		if(v==u)
    		    break;
    	}
    }
}
void find(int l,int r)
{
	int i;
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	memset(instack,0,sizeof(instack));
	memset(iscut,0,sizeof(iscut));
	dfsclock=bcccnt=0;
	for(i=l;i<=r;i++)
	{
		if(!dfn[i])
		    tarjan(i,-1);
	}
}
void suodian()
{
	int u,v,i;
	memset(in,0,sizeof(in));
	for(i=0;i<ans;i+=2)
	{
		u=bccno[edge[i].beg];
		v=bccno[edge[i].end];
		if(v!=u)
		{
			newmap[u].push_back(v);
			newmap[v].push_back(u);
			in[u]++;
			in[v]++;
		}
	}
}
void solve()
{
	int sum,i,j;
	sum=0;
	for(i=1;i<=bcccnt;i++)
	{
		if(in[i]==1)
			sum++;
	}
	printf("%d\n",(sum+1)/2);
}
int main()
{
	int i,j,k,t;
	k=1;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		getmap();
		find(1,n);
		suodian();
		solve();
	}
	return 0;
}