【BZOJ1718】&&【POJ3177】Redundant Paths

1718: [Usaco2006 Jan] Redundant Paths 分离的路径

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 452  Solved: 239
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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 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.

    为了从F(1≤F≤5000)个草场中的一个走到另一个，贝茜和她的同伴们有时不得不路过一些她们讨厌的可怕的树．奶牛们已经厌倦了被迫走某一条路，所以她们想建一些新路，使每一对草场之间都会至少有两条相互分离的路径，这样她们就有多一些选择．

    每对草场之间已经有至少一条路径．给出所有R(F-1≤R≤10000)条双向路的描述，每条路连接了两个不同的草场，请计算最少的新建道路的数量, 路径由若干道路首尾相连而成．两条路径相互分离，是指两条路径没有一条重合的道路．但是，两条分离的路径上可以有一些相同的草场． 对于同一对草场之间，可能已经有两条不同的道路，你也可以在它们之间再建一条道路，作为另一条不同的道路．

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.

    第1行输入F和R，接下来R行，每行输入两个整数，表示两个草场，它们之间有一条道路．

 

Output

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

    最少的需要新建的道路数．

Sample Input

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

Sample Output

2

HINT

 

 

Source

Gold

 

Sol：

我来详细的说一下这道题

首先 我们不难发现已经是边双联通分量的点集一定没有用了，缩缩缩。

缩完之后 我们发现我们缺少些什么？ 我们缺少要将整张图缩成一个点 

那怎么连最优？

假设我们全都向根连，肯定不优。

我也不知道怎么yy的，反正度为1的点之间互相连是最优解

 

 

（感谢caidongxiao大神的题解）

具体的证明……我们考虑怎样让度数为1的点之间互相连通就好了，肯定是互相连，但是直接随便连是有问题的，所以说应该让LCA最远的互相连一下，然后次远的相连，我也不会证明，反正是个贪心。设缩点后叶节点的个数为$leaf$个

答案就是$\lceil \frac{leaf}{2} \rceil$

边双联通分量

挺好写的 直接写

1A 基本上就是普通的tarjan加个条件

/*To The End Of The Galaxy*/
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iomanip>
#include<bitset>
#include<stack>
#include<map>
#include<set>
#include<cmath>
#include<complex>
#define debug(x) cerr<<#x<<"="<<x<<endl
#define INF 0x7f7f7f7f
#define llINF 0x7fffffffffffll
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
inline int init()
{
    int now=0,ju=1;char c;bool flag=false;
    while(1)
    {
        c=getchar();
        if(c=='-')ju=-1;
        else if(c>='0'&&c<='9')
        {
            now=now*10+c-'0';
            flag=true;
        }
        else if(flag)return now*ju;
    }
}
inline long long llinit()
{
    long long now=0,ju=1;char c;bool flag=false;
    while(1)
    {
        c=getchar();
        if(c=='-')ju=-1;
        else if(c>='0'&&c<='9')
        {
            now=now*10+c-'0';
            flag=true;
        }
        else if(flag)return now*ju;
    }
}
struct edge
{
    int from,to,val,pre;
}Edge[20005];
int head[5005],instack[5005],dfs_time,dfn[5005],low[5005];
int eccnumber,bel[5005];
int du[5005];
stack<int> s;
int topt,n,m,cnt;
map<pii,int> mp;
inline void addedge(int from,int to,int val)
{
    ++cnt;
    Edge[cnt]=((edge){from,to,val,head[from]});
    head[from]=cnt;
}
void tarjan(int now,int fa)
{
    dfn[now]=low[now]=++dfs_time;
    s.push(now);instack[now]=1;
    for(int j=head[now];j;j=Edge[j].pre)
    {
        if(!dfn[Edge[j].to])
        {
            tarjan(Edge[j].to,now);
            low[now]=min(low[now],low[Edge[j].to]);
        }
        else if(instack[Edge[j].to]&&fa!=Edge[j].to)
        {
            low[now]=min(low[now],dfn[Edge[j].to]);
        }
    }
    if(dfn[now]==low[now])
    {
        ++eccnumber;
        while(1)
        {
            topt=s.top();s.pop();
            instack[topt]=false;
            bel[topt]=eccnumber;
            if(topt==now)break;
        }
    }
}
inline void rebuild()
{
    for(int i=1;i<=cnt;i+=2)
    {
        if(bel[Edge[i].from]!=bel[Edge[i].to]&&!mp[make_pair(bel[Edge[i].from],bel[Edge[i].to])])
        {
            mp[make_pair(bel[Edge[i].from],bel[Edge[i].to])]=1;
            du[bel[Edge[i].from]]++;du[bel[Edge[i].to]]++;
        }
    }
}
int main()
{
    int a,b;
    n=init();m=init();
    for(int i=1;i<=m;i++)
    {
        a=init();b=init();
        addedge(a,b,1);
        addedge(b,a,1);
    }
    for(int i=1;i<=n;i++)
    {
        if(!dfn[i])
        {
            tarjan(i,0);
        }
    }
    rebuild();
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        if(du[i]==1)
        {
            ans++;
        }
    }
    ans++;
    printf("%d\n",ans/2);
    return 0;
}