【JZOJ 5394】【NOIP2017提高A组模拟10.5】Ping

Description

Solution

考虑链的情况，有一个显然的贪心，按左边的的排序，那么最右边的左边时一定选的，依次类推，

把结论推到树上，发现是以LCA的深度来排序的，证明显然，

复杂度： O(nlog(n))

Code

#include 
#include 
#include 
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
#define efo(i,q) for(int i=A[q];i;i=B[i][0])
#define min(q,w) ((q)>(w)?(w):(q))
#define max(q,w) ((q)<(w)?(w):(q))
#define NX(q) ((q)&(-(q)))
using namespace std;
const int N=100500;
int read(int &n)
{
    char ch=' ';int q=0,w=1;
    for(;(ch!='-')&&((ch<'0')||(ch>'9'));ch=getchar());
    if(ch=='-')w=-1,ch=getchar();
    for(;ch>='0' && ch<='9';ch=getchar())q=q*10+ch-48;n=q*w;return n;
}
int m,n,ans;
int B[N*2][2],A[N],B0;
int B1[N*3][3],A1[N],B10;
int sum[N],dfn;
int Ans[N];
struct qqww
{
    int c,i,si;
    int g[20];
}a[N];
void link(int q,int w)
{
    B[++B0][0]=A[q],A[q]=B0,B[B0][1]=w;
    B[++B0][0]=A[w],A[w]=B0,B[B0][1]=q;
}
void link1(int q,int w,int w1)
{
    B1[++B10][0]=A1[q],A1[q]=B10,B1[B10][1]=w,B1[B10][2]=w1;
}
int dfsf(int q,int fa,int c)
{
    a[q].c=c;a[q].g[0]=fa;a[q].i=++dfn;
    efo(i,q)if(B[i][1]!=fa)a[q].si+=dfsf(B[i][1],q,c+1);
    return ++a[q].si;
}
int LCA1(int q,int w)
{
    while(a[q].c>w)
    {
        int i=0;
        for(;a[a[q].g[i+1]].c>=w;i++);
        q=a[q].g[i];
    }
    return q;
}
int LCA(int q,int w)
{
    q=LCA1(q,a[w].c);w=LCA1(w,a[q].c);
    while(q!=w)
    {
        int i=0;
        for(;a[q].g[i+1]!=a[w].g[i+1];i++);
        q=a[q].g[i];w=a[w].g[i];
    }
    return q;
}
int Gsum(int q)
{
    int ans=0;
    for(;q;q-=NX(q))ans+=sum[q];
    return ans;
}
void ADD(int q,int e){for(;q<=n;q+=NX(q))sum[q]+=e;}
void dfs(int q,int fa)
{
    efo(i,q)if(B[i][1]!=fa)
    {
        dfs(B[i][1],q);
    }
    for(int i=A1[q];i;i=B1[i][0])
    {
        if(!Gsum(a[B1[i][1]].i)&&!Gsum(a[B1[i][2]].i))
        {
            Ans[++ans]=q;
            ADD(a[q].i,1);
            ADD(a[q].i+a[q].si,-1);
            break;
        }
    }
}
int main()
{
    freopen("ping.in","r",stdin);
    freopen("ping.out","w",stdout);
    int q,w,e;
    read(n),read(m);
    fo(i,1,m)read(q),read(w),link(q,w);
    fo(i,1,n)if(!a[i].g[0])dfsf(i,0,1);
    fo(j,1,18)fo(i,1,n)a[i].g[j]=a[a[i].g[j-1]].g[j-1];
    read(m);
    fo(i,1,m)
    {
        read(q),read(w);
        e=LCA(q,w);
        if(e)link1(e,q,w);
    }
    fo(i,1,n)if(!a[i].g[0])dfs(i,0);
    printf("%d\n",ans);
    fo(i,1,ans)printf("%d ",Ans[i]);
    return 0;
}