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("\n"); } return 0; }