poj3177Redundant Paths【构造双连通分量：并查集缩点 模板】

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

这个题就是问加几条边可以构成双连通分量，一开始图样图森破的以为只是求桥的个数就好，然而并非如此……

构造双连通分量的加边数=（原图的叶节点数+1）/2    因为双连通分量需要成环嘛，原图已经是连着的了，所以只需要在另一侧再加一条边就成环啦~怎么判断哪里是叶结点呢？先用并查集缩点，把所有当前的双连通分量都缩到一起，然后就构成了只有桥的图，枚举每个桥，记录每个点的次数，每次加一。只有1的点就是原图的叶结点~。~

而且并查集不就是干这个用的么╮(╯_╰)╭

/***************
poj3177
2015.11.20
***************/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>

using namespace std;

const int N=5006;
vector<int>G[N];
struct bridge
{
    int u,v;
}bg[2*N];
int vis[N],low[N],dfn[N],Time;
int fa[N],deg[N];
int n,m,cnt;
void init()
{
    for(int i=0;i<n;i++) G[i].clear();
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(vis,0,sizeof(vis));
    memset(deg,0,sizeof(deg));
    for(int i=1;i<=n;i++) fa[i]=i;
    cnt=Time=0;
}
int findset(int x)
{
    if(x!=fa[x])
        fa[x]=findset(fa[x]);
    return fa[x];
}
void Tarjan(int u,int father)
{
    low[u] = dfn[u] = ++Time;
    vis[u] = 1;
    for(int i=0;i<G[u].size();i++)
    {
        int v = G[u][i];
        if(v == father)
            continue;
        if(!vis[v])
        {
            Tarjan(v,u);
            low[u] = min(low[u],low[v]);
            if(low[v] > dfn[u])        //u->v为桥
                bg[cnt].u = u,bg[cnt++].v = v;
            else   //否则，u,v同属一个连通分量，合并
            {
                int fx = findset(u);
                int fy = findset(v);
                if(fx != fy)
                    fa[fx] = fy;
            }
        }
        else
            low[u] = min(low[u],dfn[v]);
    }
}

int main()
{
   // freopen("cin.txt","r",stdin);
    while(~scanf("%d%d", &n, &m))
    {
        init();
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        Tarjan(1,-1);
        for(int i=0;i<cnt;i++)
        {
            int fx=findset(bg[i].u);
            int fy=findset(bg[i].v);
            deg[fx]++;
            deg[fy]++;
        }
        int leaf=0;
        for(int i=1;i<=n;i++)
            if(deg[i]==1)
                leaf++;
        printf("%d\n",(leaf+1)/2);
    }
    return 0;

}