Codeforces Round #375 (Div. 2) F. st-Spanning Tree 生成树

F. st-Spanning Tree

题目连接：

http://codeforces.com/contest/723/problem/F

Description

You are given an undirected connected graph consisting of n vertices and m edges. There are no loops and no multiple edges in the graph.

You are also given two distinct vertices s and t, and two values ds and dt. Your task is to build any spanning tree of the given graph (note that the graph is not weighted), such that the degree of the vertex s doesn't exceed ds, and the degree of the vertex t doesn't exceed dt, or determine, that there is no such spanning tree.

The spanning tree of the graph G is a subgraph which is a tree and contains all vertices of the graph G. In other words, it is a connected graph which contains n - 1 edges and can be obtained by removing some of the edges from G.

The degree of a vertex is the number of edges incident to this vertex.

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 200 000, 1 ≤ m ≤ min(400 000, n·(n - 1) / 2)) — the number of vertices and the number of edges in the graph.

The next m lines contain the descriptions of the graph's edges. Each of the lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) — the ends of the corresponding edge. It is guaranteed that the graph contains no loops and no multiple edges and that it is connected.

The last line contains four integers s, t, ds, dt (1 ≤ s, t ≤ n, s ≠ t, 1 ≤ ds, dt ≤ n - 1).

Output

If the answer doesn't exist print "No" (without quotes) in the only line of the output.

Otherwise, in the first line print "Yes" (without quotes). In the each of the next (n - 1) lines print two integers — the description of the edges of the spanning tree. Each of the edges of the spanning tree must be printed exactly once.

You can output edges in any order. You can output the ends of each edge in any order.

If there are several solutions, print any of them.

Sample Input

3 3

1 2

2 3

3 1

1 2 1 1

Sample Output

Yes

3 2

1 3

Hint

题意

给你一个无向图，问你能不能找到一颗生成树，使得这个生成树包含S点和T点，且S点的度数不超过DS，T点的度数不超过DT

题解：

我们首先把S点和T点都拿走，然后跑一个生成树，那么现在的图就是一个森林了。

首先我们让S和T都连到同一个连通块去。

然后再贪心的去连，S优先连T不能够连接的连通块，再让T连接S不能连接的连通块，再去连接都能够连接的连通块。

其实感觉上是一个乱搞题[二哈]

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
vector<pair<int,int> >ans;
vector<int>E[maxn];
int s,t,ds,dt,fa[maxn],vis[maxn],n,m,link1[maxn],link2[maxn];
vector<int> c;
int fi(int x)
{
    return fa[x]==x?x:fa[x]=fi(fa[x]);
}
void uni(int x,int y)
{
    x=fi(x),y=fi(y);
    fa[x]=y;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        E[x].push_back(y);
        E[y].push_back(x);
    }
    for(int i=1;i<=n;i++)
        fa[i]=i;
    scanf("%d%d%d%d",&s,&t,&ds,&dt);
    for(int i=1;i<=n;i++)
    {
        if(i==s||i==t)continue;
        for(int j=0;j<E[i].size();j++)
        {
            int v=E[i][j];
            if(v==s||v==t)continue;
            if(fi(i)!=fi(v))
            {
                ans.push_back(make_pair(i,v));
                uni(i,v);
            }
        }
    }
    int flag = 0;
    for(int i=0;i<E[s].size();i++)
    {
        int v=E[s][i];
        link1[fi(E[s][i])]=1;
    }
    for(int i=0;i<E[t].size();i++)
    {
        int v=E[t][i];
        link2[fi(E[t][i])]=1;
    }
    if(!flag)
    {
        for(int i=0;i<E[t].size();i++)
        {
            int v=E[t][i];
            if(link1[fi(v)])
            {
                vis[fi(v)]=1;
                dt--;
                ans.push_back(make_pair(t,v));
                uni(t,v);
                break;
            }
        }
        for(int i=0;i<E[s].size();i++)
        {
            int v=E[s][i];
            if(vis[v])
            {
                ds--;
                ans.push_back(make_pair(s,v));
                uni(s,v);
                break;
            }
        }
    }
    for(int i=0;i<E[s].size();i++)
    {
        if(!link2[E[s][i]]&&ds&&fi(s)!=fi(E[s][i]))
        {
            ans.push_back(make_pair(s,E[s][i]));
            uni(s,E[s][i]);
            ds--;
        }
    }
    for(int i=0;i<E[t].size();i++)
    {
        if(!link1[E[t][i]]&&dt&&fi(t)!=fi(E[t][i]))
        {
            ans.push_back(make_pair(t,E[t][i]));
            uni(t,E[t][i]);
            dt--;
        }
    }
    for(int i=0;i<E[s].size();i++)
    {
        if(ds&&fi(s)!=fi(E[s][i]))
        {
            ans.push_back(make_pair(s,E[s][i]));
            uni(s,E[s][i]);
            ds--;
        }
    }
    for(int i=0;i<E[t].size();i++)
    {
        if(dt&&fi(t)!=fi(E[t][i]))
        {
            ans.push_back(make_pair(t,E[t][i]));
            uni(t,E[t][i]);
            dt--;
        }
    }
    if(fi(s)!=fi(t))
    {
        for(int i=0;i<E[s].size();i++)
        {
            if(E[s][i]==t)
            {
                uni(s,t);
                ans.push_back(make_pair(s,t));
            }
        }
    }
    if(ans.size()==n-1)
    {
        cout<<"Yes"<<endl;
        for(int i=0;i<ans.size();i++)
            cout<<ans[i].first<<" "<<ans[i].second<<endl;
    }
    else
        cout<<"No"<<endl;
}