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  • bzoj2588

    题目:http://www.lydsy.com/JudgeOnline/problem.php?id=2588

    就是静态区间第K大的变形。
    每个节点为一棵线段树,表示到根的路径中,以权值为下标的线段树。
    每个节点建树的时候,以父亲为历史版本。
    对于询问点x和点y的时候,就是求ask(x)+ask(y)-ask(lca)-ask(fa[lca]),然后用类似于静态区间第K的方法决定是去左子树还是右子树。
    #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 re(i,a,b)  for(i=a;i<=b;i++)
    #define red(i,a,b) for(i=a;i>=b;i--)
    #define fi first
    #define se second
    #define m_p(a,b) make_pair(a,b)
    #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;}
    
    const DB EPS=1e-9;
    inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
    const DB Pi=acos(-1.0);
    
    inline 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; 
        }
    inline 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,M;
    int a[maxN+100];
    int bak[maxN+100],cnt;
    int first[maxN+100],now;
    struct Tedge{int v,next;}edge[2*maxN+100];
    
    inline void addedge(int u,int v){now++;edge[now].v=v;edge[now].next=first[u];first[u]=now;}
    
    int head,tail,que[maxN+100];
    int fa[maxN+100],dep[maxN+100];
    int jump[35][maxN+100];
    
    struct Tnode{int val,son[2];}sn[maxN*70+100000];int idx;
    
    int root[maxN+100];
    
    inline void update(int past,int p,int l,int r,int x)
      {
          if(x<=l && r<=x){sn[p].val=sn[past].val+1;return;}
          int mid=(l+r)/2,f=(x<=mid);
          sn[p].son[f]=sn[past].son[f];
          sn[p].son[f^1]=++idx;
          if(x<=mid) update(sn[past].son[0],sn[p].son[0],l,mid,x); else update(sn[past].son[1],sn[p].son[1],mid+1,r,x);
          sn[p].val=sn[sn[p].son[0]].val+sn[sn[p].son[1]].val;
      }
    inline int ask(int r1,int r2,int r3,int r4,int l,int r,int k)
      {
          if(l==r)return bak[l];
          int mid=(l+r)/2;
          int ge=sn[sn[r1].son[0]].val+sn[sn[r2].son[0]].val-sn[sn[r3].son[0]].val-sn[sn[r4].son[0]].val;
          if(ge<k)
            return ask(sn[r1].son[1],sn[r2].son[1],sn[r3].son[1],sn[r4].son[1],mid+1,r,k-ge);
          else
            return ask(sn[r1].son[0],sn[r2].son[0],sn[r3].son[0],sn[r4].son[0],l,mid,k);
      }
    
    inline void BFS()
      {
          int i,j;
          mmst(fa,-1);
          fa[que[head=tail=0]=1]=0;
          dep[1]=1;
          while(head<=tail)
            {
                int u=que[head++],v;
                for(i=first[u],v=edge[i].v;i!=-1;i=edge[i].next,v=edge[i].v)if(fa[v]==-1){fa[que[++tail]=v]=u;dep[v]=dep[u]+1;}
            }
          re(i,0,N)root[i]=++idx;
          re(i,0,tail)
              {
                  int u=que[i];
                    update(root[fa[u]],root[u],1,cnt,a[u]);
                }
          re(j,0,30)jump[j][1]=1;
          re(i,1,tail)
            {
                int u=que[i];
                jump[0][u]=fa[u];
                re(j,1,30)jump[j][u]=jump[j-1][jump[j-1][u]];
            }
      }
    
    inline void swim(int &x,int H){int i;for(i=0;H!=0;H/=2,i++)if(H&1)x=jump[i][x];}
    inline int ask_lca(int x,int y)
      {
          if(dep[x]<dep[y])swap(x,y);
          swim(x,dep[x]-dep[y]);
          if(x==y)return x;
          int i;
          red(i,30,0)if(jump[i][x]!=jump[i][y]){x=jump[i][x];y=jump[i][y];}
          return jump[0][x];
      }
    
    int main()
      {
          freopen("bzoj2588.in","r",stdin);
          freopen("bzoj2588.out","w",stdout);
          int i;
          N=gint();M=gint();
          re(i,1,N)a[i]=gint();
          re(i,1,N)bak[i]=a[i];
          sort(bak+1,bak+N+1);
          cnt=unique(bak+1,bak+N+1)-bak-1;
          re(i,1,N)a[i]=lower_bound(bak+1,bak+cnt+1,a[i])-bak;
          mmst(first,-1);now=-1;
          re(i,1,N-1){int x=gint(),y=gint();addedge(x,y);addedge(y,x);}
          BFS();
          int lastans=0;
          while(M--)
            {
                int x=gint()^lastans,y=gint(),lca=ask_lca(x,y),k=gint();
                lastans=ask(root[x],root[y],root[lca],root[fa[lca]],1,cnt,k);
                printf("%d",lastans);if(M!=0)printf("
    ");
            }
          return 0;
      }
    View Code
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  • 原文地址:https://www.cnblogs.com/maijing/p/4649480.html
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