POJ-3177 Redundant Paths （边双连通分量[Tarjan]）

Redundant Paths

http://poj.org/problem?id=3177

Time Limit: 1000MS		Memory Limit: 65536K

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.

题目大意：给定一个无向图，求加入最少的边，使该图变成一个双连通分量？

点双连通分量：不存在割点的连通分量

边双连通分量：不存在桥的连通分量（即任意两点互相可达的路径有不同的两条）

大致方法：可以求出所有的桥，把桥删掉。然后把所有的边双连通分量求出来，显然这些边双连通分量就是原图中的双连通分量。把它们缩成点，然后添上刚才删去的桥，就构成了一棵树。在树上添边使得树变成一个双连通分量即可。可以统计该树度为1的叶子节点（设共有x个），则答案为：(x+1)/2

注意：题目数据会给出重边，但是重边只算一次（即题目意思是两点之间有路，这个信息可能会重复出现，但实际上只有一条路），但是我看有一个题解没有判断重边也过了，很是费解（这个题解用数组模拟的邻接表，判断父子边时，由于父子边的下标只差1，所以通过异或运算判断，但是不能判断重边，然后就不明白了）

附上课件中的证明：

命题：一棵有n(n>=2)个叶子结点的树，至少(只需）要添加ceil(n/2)条边，才(就)能转变为一个没有桥的图。或者说，使得图中每条边，都至少在一个环上。
证明：
这里只证明n为偶数的情况。n为奇数的证明类似。
先证明添加n/2条边一定可以达成目标。
n=2时，显然只需将这两个叶子间连一条边即可。命题成立。
设n=2k(k>=1)时命题成立，即AddNum(2k)=k。下面将推出n=2(k+1)时命题亦成立
n=2k+2时，选取树中一条迹(无重复点的路径），设其端点为a,b；并设离a最近的度>=3的点为a',同理设b'。
（关于a‘和b’的存在性问题：由于a和b的度都为1,因此树中其它的树枝必然从迹<a,b>之间的某些点引出。否则整棵树就是迹<a,b>，n=2<2k+2，不可能。）

a’ b’不重合时：
在a,b间添一条边，则迹<a,b>上的所有边都已不再是桥。这时，将刚才添加的边，以及aa‘之间，bb’之间的边都删去，得到一棵新的树。因为删去的那些边都已经符合条件了，所以在之后的构造中不需要考虑它们。由于之前a‘和b’的度>=3，所以删除操作不会使他们变成叶子。因此新的树必然比原树少了两个叶子a,b，共有2k个叶子。由归纳知需要再加k条边。因此对n=2k+2的树，一共要添加k+1条边。

将a和一个非b的叶子节点x连上，然后将环缩点至 a’。
因为叶子节点是偶数，所以必然还存在一个非b非x的叶子节点不在环上， 因此a’不会变成叶子节点，于是新图比原图少2个叶子节点。

再证明n/2是最小的解。
显然，只有一个叶子结点被新加的边覆盖到，才有可能使与它相接的那条边进入一个环中。而一次加边至多覆盖2个叶子。因此n个叶子至少要加n/2条边。
证毕。

#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

const int MAXN=5005;

int n,m,num,ans,cnt,top;
int stak[MAXN],low[MAXN],dfn[MAXN],color[MAXN],deg[MAXN];//biocks[i]表示删掉i点后，能形成的连通块数
vector<int> g[MAXN];
bool isIn[MAXN],vis[MAXN][MAXN];

void Tarjan(int u,int p) {//求无向图的边双连通分量的Tarjan与求有向图的强连通分量的Tarjan好像...
    stak[++top]=u;
    isIn[u]=true;
    dfn[u]=low[u]=++num;
    int v;
    for(int i=0;i<g[u].size();++i) {
        v=g[u][i];
        if(v!=p) {
            if(dfn[v]==0) {
                Tarjan(v,u);
                low[u]=min(low[u],low[v]);
            }
            else if(isIn[v]&&dfn[v]<low[u])
                low[u]=dfn[v];
        }
    }

    if(dfn[u]==low[u]) {
        ++cnt;
        do {
            v=stak[top--];
            isIn[v]=false;
            color[v]=cnt;//一个边双连通分量缩成一个点
        } while(v!=u);
    }
}

int solve() {
    cnt=num=top=0;//cnt表示边双连通分量的个数
    memset(dfn,0,sizeof(dfn));
    memset(isIn,false,sizeof(isIn));
    for(int i=1;i<=n;++i) {
        if(dfn[i]==0)
            Tarjan(i,0);
    }

    memset(deg,0,sizeof(deg));
    for(int i=1;i<=n;++i) {
        for(int j=0;j<g[i].size();++j)
            if(color[i]!=color[g[i][j]])//若这两个点不再同一个双连通分量中
                ++deg[color[i]];//由于是无向图，所以只对起点的度+1即可
    }
    int ans=0;
    for(int i=1;i<=cnt;++i)
        if(deg[i]==1)
            ++ans;
    return (ans+1)>>1;
}

int main() {
    int s,e;
    while(scanf("%d%d",&n,&m)==2) {
        for(int i=1;i<=n;++i) {
            g[i].clear();
            for(int j=i;j<=n;++j)
                vis[i][j]=vis[j][i]=false;
        }

        while(m-->0) {
            scanf("%d%d",&s,&e);
            if(!vis[s][e]) {//去掉重边
                vis[s][e]=vis[e][s]=true;
                g[s].push_back(e);
                g[e].push_back(s);
            }
        }
        printf("%d\n",solve());
    }
    return 0;
}