• bzoj1095[ZJOI2007]Hide 捉迷藏


    http://www.lydsy.com/JudgeOnline/problem.php?id=1095

    好像有2种做法:线段树维护括号编码&动态树分治。

    线段树维护括号编码:

    《数据结构的提炼与压缩》

    这篇论文是讲得极好的。

    #include<cstdio>
    #include<cstdlib>
    #include<iostream>
    #include<fstream>
    #include<algorithm>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<queue>
    #include<stack>
    #include<map>
    #include<utility>
    #include<set>
    #include<bitset>
    #include<vector>
    #include<functional>
    #include<deque>
    #include<cctype>
    #include<climits>
    #include<complex>
    //#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj
     
    using namespace std;
    
    typedef long long LL;
    typedef double DB;
    typedef pair<int,int> PII;
    typedef complex<DB> CP;
    
    #define mmst(a,v) memset(a,v,sizeof(a))
    #define mmcy(a,b) memcpy(a,b,sizeof(a))
    #define fill(a,l,r,v) fill(a+l,a+r+1,v)
    #define re(i,a,b)  for(i=(a);i<=(b);i++)
    #define red(i,a,b) for(i=(a);i>=(b);i--)
    #define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)
    #define fi first
    #define se second
    #define m_p(a,b) make_pair(a,b)
    #define p_b(a) push_back(a)
    #define SF scanf
    #define PF printf
    #define two(k) (1<<(k))
    
    template<class T>inline T sqr(T x){return x*x;}
    template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
    template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}
    
    inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;}
    const DB Pi=acos(-1.0);
    
    int gint()
      {
            int res=0;bool neg=0;char z;
            for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
            if(z==EOF)return 0;
            if(z=='-'){neg=1;z=getchar();}
            for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
            return (neg)?-res:res; 
        }
    LL gll()
      {
          LL res=0;bool neg=0;char z;
            for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
            if(z==EOF)return 0;
            if(z=='-'){neg=1;z=getchar();}
            for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
            return (neg)?-res:res; 
        }
    
    const int maxn=100000;
    
    int n;
    int now,info[maxn+100];
    struct Tedge{int v,next;}edge[2*maxn+100];
    int status[maxn+100],ge;
    
    void addedge(int u,int v){now++;edge[now].v=v;edge[now].next=info[u];info[u]=now;}
    
    int cnt,ty[3*maxn+100],pos[maxn+100];
    void dfs(int u)
      {
          ty[++cnt]=-1;
          ty[++cnt]=u;
          pos[u]=cnt;
          int i,v;
          for(i=info[u],v=edge[i].v;i!=-1;i=edge[i].next,v=edge[i].v)if(!pos[v])dfs(v);
          ty[++cnt]=-2;
      }
    
    struct Tdata
      {
          int a,b;
          Tdata(int _a=0,int _b=0){a=_a;b=_b;}
          friend Tdata operator +(Tdata l,Tdata r){return (l.b<=r.a)?Tdata(l.a+r.a-l.b,r.b):Tdata(l.a,l.b-r.a+r.b);}
          friend bool operator ==(Tdata x,Tdata y){return x.a==y.a && x.b==y.b;}
          friend bool operator !=(Tdata x,Tdata y){return !(x==y);}
      };
    Tdata null=Tdata(-1,-1);
    
    /*
    dist 子区间的最大值 
    whole 区间[l,r]的二元组(a,b) 
    lp    右端点为黑点的a+b最大的前缀的二元组(a,b)
    rp    左端点为黑点的a+b最大的后缀的二元组(a,b)
    lm    右端点为黑点的b-a最大的前缀的二元组(a,b)
    rm    左端点为黑点的a-b最大的后缀的二元组(a,b)
    null表示不存在 
    */  
    struct Tnode
      {
          int dist;
          Tdata whole,lp,rp,lm,rm;
          void init(int x);
      };
    
    void Tnode::init(int x)
      {
          int lcnt=(ty[x]==-2),rcnt=(ty[x]==-1);
          whole=Tdata(lcnt,rcnt);
          dist=-1;
          if(ty[x]>=1 && status[ty[x]]) lp=rp=lm=rm=whole; else lp=rp=lm=rm=null;
      }
    
    Tnode make_up(Tnode x,Tnode y)
      {
          Tnode res;Tdata p;
          
          res.whole=x.whole+y.whole;
          
          res.lp=x.lp;
          if(y.lp!=null)
            {
                p=x.whole+y.lp;
                if(res.lp==null || res.lp.a+res.lp.b<p.a+p.b)res.lp=p;
            }
          if(y.lm!=null)
            {
                p=x.whole+y.lm;
                if(res.lp==null || res.lp.a+res.lp.b<p.a+p.b)res.lp=p;
            }
          
          res.rp=y.rp;
          if(x.rp!=null)
            {
                p=x.rp+y.whole;
                if(res.rp==null || res.rp.a+res.rp.b<p.a+p.b)res.rp=p;
            }
          if(x.rm!=null)
            {
                p=x.rm+y.whole;
                if(res.rp==null || res.rp.a+res.rp.b<p.a+p.b)res.rp=p;
            }
          
          res.lm=x.lm;
          if(y.lm!=null)
            {
                p=x.whole+y.lm;
                if(res.lm==null || res.lm.b-res.lm.a<p.b-p.a)res.lm=p;
            }
          
          res.rm=y.rm;
          if(x.rm!=null)
            {
                p=x.rm+y.whole;
                if(res.rm==null || res.rm.a-res.rm.b<p.a-p.b)res.rm=p;
            }
          
          res.dist=max(x.dist,y.dist);
          if(x.rp!=null && y.lm!=null)
              {
                  p=x.rp+y.lm;
                    upmax(res.dist,p.a+p.b);
                }
          if(x.rm!=null && y.lp!=null)
              {
                  p=x.rm+y.lp;
                    upmax(res.dist,p.a+p.b);
                }
          
          return res;
          
      }
    
    Tnode tr[4*3*maxn+100];
    void build(int rt,int l,int r)
      {
          if(l==r){tr[rt].init(l);return;}
          int mid=(l+r)>>1;
          build(rt<<1,l,mid);
          build(rt<<1|1,mid+1,r);
          tr[rt]=make_up(tr[rt<<1],tr[rt<<1|1]);
      }
    void update(int rt,int l,int r,int x)
      {
          if(l==r){tr[rt].init(l);return;}
          int mid=(l+r)/2;
          if(x<=mid)update(rt<<1,l,mid,x);else update(rt<<1|1,mid+1,r,x);
          tr[rt]=make_up(tr[rt<<1],tr[rt<<1|1]);
      }
    
    int main()
      {
          freopen("bzoj1095.in","r",stdin);
          freopen("bzoj1095.out","w",stdout);
          int i;
          n=gint();
          now=-1;mmst(info,-1);
          re(i,1,n-1){int u=gint(),v=gint();addedge(u,v);addedge(v,u);}
          dfs(1);
            re(i,1,n)status[i]=1;ge=n;
          build(1,1,cnt);
          int Q;SF("%d
    ",&Q);
          while(Q--)
            {
                char t;SF("%c",&t);
                if(t=='G')
                  {
                      if(ge==0)PF("-1
    ");else if(ge==1) PF("0
    "); else PF("%d
    ",tr[1].dist);
                  }
                else
                  {
                          int x;SF("%d",&x);
                            status[x]^=1;
                            if(status[x])ge++;else ge--;
                            update(1,1,cnt,pos[x]);
                        }
                    SF("
    ");
            }
          return 0;
      }
    View Code

    动态树分治:

    首先,我们想平常树分治那样,找到一个重心。

    然后将这个重心作为新的根,将它的所有儿子及其子树划分成大小相等的两个整体。(普通的树分治是一个儿子作为一个整体,但是这样合并的时候有点麻烦,如果只分成两个整体,合并就比较简单)

    分别用一个堆维护这两个整体的黑点到重心的距离。

    那么经过重心的最长的路径就是这两个堆的最大值之和。

    递归处理这两个整体。

    每个点最多属于$logn$个重心。

    看程序比较好理解。

    #include<cstdio>
    #include<cstdlib>
    #include<iostream>
    #include<fstream>
    #include<algorithm>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<queue>
    #include<stack>
    #include<map>
    #include<utility>
    #include<set>
    #include<bitset>
    #include<vector>
    #include<functional>
    #include<deque>
    #include<cctype>
    #include<climits>
    #include<complex>
    //#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj
     
    using namespace std;
    
    typedef long long LL;
    typedef double DB;
    typedef pair<int,int> PII;
    typedef complex<DB> CP;
    
    #define mmst(a,v) memset(a,v,sizeof(a))
    #define mmcy(a,b) memcpy(a,b,sizeof(a))
    #define fill(a,l,r,v) fill(a+l,a+r+1,v)
    #define re(i,a,b)  for(i=(a);i<=(b);i++)
    #define red(i,a,b) for(i=(a);i>=(b);i--)
    #define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)
    #define fi first
    #define se second
    #define m_p(a,b) make_pair(a,b)
    #define p_b(a) push_back(a)
    #define SF scanf
    #define PF printf
    #define two(k) (1<<(k))
    
    template<class T>inline T sqr(T x){return x*x;}
    template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
    template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}
    
    inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;}
    const DB Pi=acos(-1.0);
    
    int gint()
      {
            int res=0;bool neg=0;char z;
            for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
            if(z==EOF)return 0;
            if(z=='-'){neg=1;z=getchar();}
            for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
            return (neg)?-res:res; 
        }
    LL gll()
      {
          LL res=0;bool neg=0;char z;
            for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
            if(z==EOF)return 0;
            if(z=='-'){neg=1;z=getchar();}
            for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
            return (neg)?-res:res; 
        }
    
    const int maxn=100000;
    const int inf=0x3f3f3f3f;
    
    struct Priority_Queue
      {
          priority_queue<int>heap,delmark;
          void push(int x){heap.push(x);}
          void erase(int x){delmark.push(x);}
          void clear(){while(!delmark.empty() && !heap.empty() && delmark.top()==heap.top())heap.pop(),delmark.pop();}
          int top(){clear();return heap.empty()?-inf:heap.top();}
          void pop(){clear();heap.pop();}
          bool empty(){return heap.size()==delmark.size();}
      };
    
    int n;
    int now[2],info[2][maxn+100];
    struct Tedge{int v,next,kk,dis;}edge[2][3000000];//注意边数是nlogn 
    int status[maxn+100];
    
    void addedge(int k,int u,int v,int kk=0,int dis=0)
      {
          now[k]++;
          edge[k][now[k]].v=v;
          edge[k][now[k]].next=info[k][u];
          if(k){edge[k][now[k]].kk=kk;edge[k][now[k]].dis=dis;}
          info[k][u]=now[k];
      }
    
    int num;
    PII cd[maxn+100];
    Priority_Queue Q[maxn+100][2];
    int ans[maxn+100];
    
    int sz[maxn+100],mx[maxn+100];
    void getroot(int u,int fa,int all,int &rt)
      {
          sz[u]=1;mx[u]=0;
          int i,v;
          for(i=info[0][u],v=edge[0][i].v;i!=-1;i=edge[0][i].next,v=edge[0][i].v)if(v!=fa)
            {
                getroot(v,u,all,rt);
                sz[u]+=sz[v];
                upmax(mx[u],sz[v]);
            }
          upmax(mx[u],all-sz[u]);
          if(!rt || mx[u]<mx[rt])rt=u;
      }
    
    void dfs(int u,int fa,int dis,int cur,int kind)
      {
          addedge(1,u,cur,kind,dis);
          Q[cur][kind].push(dis);
          int i,v;
          for(i=info[0][u],v=edge[0][i].v;i!=-1;i=edge[0][i].next,v=edge[0][i].v)if(v!=fa)dfs(v,u,dis+1,cur,kind);
      }
    
    void calc(int cur)
      {
          ans[cur]=max(ans[cd[cur].fi],ans[cd[cur].se]);
          if(!Q[cur][0].empty() && !Q[cur][1].empty())upmax(ans[cur],Q[cur][0].top()+Q[cur][1].top());
      }
    
    int prep(int u,int all)
      {
          if(all<=2)return 0;
          int i,rt=0,tmp=0;
          getroot(u,-1,all,rt);
          getroot(rt,-1,all,tmp);
          int cur=++num,sum=0,v;
          for(i=info[0][rt],v=edge[0][i].v;i!=-1;i=edge[0][i].next,v=edge[0][i].v)
              {
                  sum+=sz[v];
                    if(sum>=(all-1)/2 || edge[0][i].next==-1)break;
                }
          int st1=info[0][rt],ed1=i,st2=edge[0][ed1].next;
          edge[0][ed1].next=-1;
          dfs(rt,-1,0,cur,0);
          cd[cur].fi=prep(rt,sum+1);
          edge[0][ed1].next=info[0][rt]=st2;
          dfs(rt,-1,0,cur,1);
          cd[cur].se=prep(rt,all-sum);
          info[0][rt]=st1;
          calc(cur);
          return cur;
          
      }
    
    void update(int x)
      {
          int i,cur,dis,kind;
          for(i=info[1][x],cur=edge[1][i].v,dis=edge[1][i].dis,kind=edge[1][i].kk;i!=-1;i=edge[1][i].next,cur=edge[1][i].v,dis=edge[1][i].dis,kind=edge[1][i].kk)
            {
                if(status[x])Q[cur][kind].push(dis);else Q[cur][kind].erase(dis);
                calc(cur);
            }
      }
    
    int main()
      {
          freopen("bzoj1095.in","r",stdin);
          freopen("bzoj1095.out","w",stdout);
          int i;
          n=gint();
          mmst(now,-1);mmst(info,-1);
          re(i,1,n-1){int u=gint(),v=gint();addedge(0,u,v);addedge(0,v,u);}
          re(i,1,n)status[i]=1;
          ans[0]=-inf;
          prep(1,n);
          int ge=n,T;SF("%d
    ",&T);
          while(T--)
            {
                char z=getchar();
                if(z=='G')
                  {
                      if(ge==0)PF("-1
    ");else if(ge==1) PF("0
    ");else PF("%d
    ",ans[1]);
                  }
                else
                  {
                      int x;SF("%d",&x);
                      status[x]^=1;
                      if(status[x])ge++;else ge--;
                      update(x);
                  }
                SF("
    ");
            }
          return 0;
      }
    View Code
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  • 原文地址:https://www.cnblogs.com/maijing/p/4970245.html
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