[SPOJ-COT]Count on a tree

题面

You are given a tree with N nodes. The tree nodes are numbered from 1 to N. Each node has an integer weight.

We will ask you to perform the following operation:

• u v k : ask for the kth minimum weight on the path from node u to node v

Input

In the first line there are two integers N and M. (N, M <= 100000)

In the second line there are N integers. The ith integer denotes the weight of the ith node.

In the next N-1 lines, each line contains two integers u v, which describes an edge (u, v).

In the next M lines, each line contains three integers u v k, which means an operation asking for the kth minimum weight on the path from node u to node v.

Output

For each operation, print its result.

Example

Input:
8 5
105 2 9 3 8 5 7 7
1 2
1 3
1 4
3 5
3 6
3 7
4 8
2 5 1
2 5 2
2 5 3
2 5 4
7 8 2
Output:
2
8
9
105
7 

思路

这是一个把主席树推广到树上的题。我们首先思考一下，普通的主席树/权值线段树只能对线性的区间进行操作，我们要做的，就是把一棵树分解成一个线性的结构，同时便于进行题目规定的查询操作。

因为这道题是求路径到LCA上的k小值。我们就可以把每个节点到根节点的路径建一棵树。每个节点的子节点对应的线段树可以从父节点线段树继承得到。复杂度\(n\log_2 n\)。

那解法就很显然了，我们先对根节点建一个权值线段树，然后按照DFS序依次遍历整棵树，将父节点继承下来，成为一颗新的树。

对于查询\((i,j,k)\)，我们只需要求出一棵包含\(i\)到\(j\)所有点权值的权值线段树，按照k小值标准查询方式查询即可。

那么我们设这棵线段树为\(T_a\)，从节点\(i\)到根节点的权值线段树为\(T_i\)。那么很显然：
\[ T_a=T_i+T_j-T_{LCA(i,j)}-T_{fa[LCA(i,j)]} \]
注意，至于为什么最后一项是\(fa[LCA(i,j)]\)，仔细想一想就能发现如果两个都是\(LCA(i,j)\)，会把\(LCA(i,j)\)多减一次。

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn (int)(1e5+1000)
int
n,m,idx,len,ls[maxn*21],rs[maxn*21],sum[maxn*21],a[maxn],sorted[maxn],head[maxn],cnt,rt[maxn],f[maxn];
struct gg{int u,v,next;
}side[maxn*2];
int getid(int x){return lower_bound(sorted+1,sorted+1+len,x)-sorted;
}
void insert(int u,int v){gg q={u,v,head[u]};side[++cnt]=q;head[u]=cnt;return;
}
int build(int l,int r){int now=++idx;if(l==r)return now;int mid=(l+r)>>1;ls[now]=build(l,mid);rs[now]=build(mid+1,r);return now;
}
int update(int last,int pos,int l,int r){int now=++idx;rs[now]=rs[last],ls[now]=ls[last],sum[now]=sum[last]+1;if(l==r){return now; }int mid=(l+r)>>1;if(pos<=mid){ls[now]=update(ls[last],pos,l,mid);}else{rs[now]=update(rs[last],pos,mid+1,r);}return now;
}
void dfs(int now,int fa){f[now]=fa;rt[now]=update(rt[fa],getid(a[now]),1,len);//testfor(int i=head[now];i;i=side[i].next){int v=side[i].v;if(v==fa)continue;dfs(v,now);}return;
}
int query(int p1,int p2,int p3,int p4,int k,int l,int r){//i,j,lca,fa[lca]int tot=sum[ls[p1]]+sum[ls[p2]]-sum[ls[p3]]-sum[ls[p4]];int mid=(l+r)>>1;if(l==r)return l;if(k<=tot){return query(ls[p1],ls[p2],ls[p3],ls[p4],k,l,mid);}else{return query(rs[p1],rs[p2],rs[p3],rs[p4],k-tot,mid+1,r);}return 0;
}
/*-----------------------*/
int fa[maxn][30],dep[maxn];
void dfs1(int u, int f) {dep[u] = dep[f] + 1, fa[u][0] = f;for (int i = 1;i <= 20;i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];for (int i = head[u];i;i = side[i].next)if (side[i].v != f) dfs1(side[i].v, u);
}
int lca(int a1, int b1) {if (dep[a1] < dep[b1]) swap(a1, b1);for (int i = 20;i >= 0;i--) if (dep[fa[a1][i]] >= dep[b1]) a1 = fa[a1][i];if (a1 == b1) return a1;for (int i = 20;i >= 0;i--) if (fa[a1][i] != fa[b1][i]) a1 =fa[a1][i],b1 = fa[b1][i];return fa[a1][0];
}/*------------------------*/
int main(){//freopen("in","r",stdin);scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%d",&a[i]);sorted[i]=a[i];}sort(sorted+1,sorted+1+n);len=unique(sorted+1,sorted+1+n)-sorted-1;for(int i=1;i<n;i++){int u,v;scanf("%d%d",&u,&v);insert(u,v);insert(v,u);}rt[0]=build(1,len);dfs(1,0);dfs1(1,0);for(int i=1;i<=m;i++){int u,v,k;scanf("%d%d%d",&u,&v,&k);int ans=query(rt[u],rt[v],rt[(lca(u,v))],rt[f[lca(u,v)]],k,1,len);printf("%d\n",sorted[ans]);}return 0;
}