CF 653E(Bear and Forgotten Tree 2-链表优化搜索)

给一张n个点的树（无重边自环），已知m条边不存在，已知节点1的度数，问这是否可能？

直接把1节点拿掉求连通块，然后几行特判就行。

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p]) 
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair 
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define MAXN (300000+10)
#define MAXM (3000000)
int n,m,k;  
int size=0,edge[MAXM*2],pre[MAXM],Next[MAXM*2];  
struct link  
{  
    int pre,Next;  
}l[MAXN];  
void del(int x)  
{  
    l[l[x].pre].Next=l[x].Next;  
    l[l[x].Next].pre=l[x].pre;    
}  
queue<int> q;  
void addedge(int u,int v)  
{  
    edge[++size]=v;  
    Next[size]=pre[u];  
    pre[u]=size;  
}  
bool b[MAXN],q_b[MAXN];  
int z[MAXN],siz_z=0;
vector<int> Ve[MAXN];  
void bfs()  
{  
    memset(b,0,sizeof(b));  
    memset(q_b,0,sizeof(q_b));  
    while (l[0].Next)  
    {  
        int now=l[0].Next,ans=1;
        Ve[siz_z].pb(now);
        del(now);  
        q.push(now);q_b[now]=1;  
        //cout<<now<<endl; //00x 
        while (!q.empty())  
        {  
            int u=q.front();q.pop();  
            for (int p=pre[u];p;p=Next[p]) b[edge[p]]=1;  
            for(int i=l[0].Next;i;i=l[i].Next)  
            {  
                if (!b[i]&&!q_b[i])  
                {  
                    Ve[siz_z].pb(i);

                    q_b[i]=1;  
                    q.push(i);//cout<<v<<endl; 
                    ans++;  
                    del(i);  
                }  
            }  
            for (int p=pre[u];p;p=Next[p]) b[edge[p]]=0;  
        }  
        z[++siz_z]=ans;  
    }  
}  
bool is_avi[MAXN]={0};
int main()
{
// freopen("E.in","r",stdin);
// freopen(".out","w",stdout);
    memset(pre,0,sizeof(pre));  
    n=read();m=read();k=read();
    int pp=0;
    for (int i=1;i<=m;i++)  
    {  
        int u,v;  
        scanf("%d%d",&u,&v);  
        if (u>v) swap(u,v);
        if (u==1) {
            is_avi[v]=1; ++pp; continue;
        } 
        addedge(u,v);addedge(v,u);  
    }  
    Fork(i,2,n) addedge(1,i),addedge(i,1);  

    for (int i=1;i<=n;i++) {l[i-1].Next=i;l[i].pre=i-1;} l[n].Next=0;  

    bfs();

    For(i,siz_z-1) { 
        int m=Ve[i].size();
        int flag=0;
        Rep(j,m) if (!is_avi[Ve[i][j]])  {
            flag=1; break;
        }
        if (flag==0) {
            puts("impossible"); return 0;
        }       
    }
    if (siz_z-1<=k&&k<=n-1-pp){
        puts("possible");
    } else {
        puts("impossible");     
    }

    return 0;   


}