| Time Limit: 129MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
Description
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
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
题意：求树上的边[u,v]中点权第k大
使用的是主席树+LCA(RMQ.dfs),然后去专门看了下RMQ+dfs实现LCA
用一个数组记录深度，然后记录搜索的路径，如果要找[a,b]中的LCA，直接找[a,b]中的深度最小值即可
参考：算法之LCA与RMQ问题
/*
主席树-代码参考kuangbin大神
在本题中相当于按树的节点来构建线段树，每个节点基于它的父亲进行构建
然后节点a保存的便是根到a的情况，于是乎我们T[a]+T[b]-2*T[lca(a,b)]即可
而且对lca节点进行一个判断。
hhh-2016-02-18 21:11:14
*/
#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
using namespace std;
const int maxn = 200010;
int n,m;
int a[maxn],t[maxn];
int T[maxn*40],val[maxn*40],lson[maxn*40],rson[maxn*40];
int Tot;
void ini_hash() //排序去重
{
for(int i =1; i <= n; i++)
t[i] = a[i];
sort(t+1,t+n+1);
m = unique(t+1,t+n+1)-t-1;
}
int Hash(int x) //获得x在排序去重后的位置
{
return lower_bound(t+1,t+m+1,x) - t;
}
int build(int l,int r)
{
int root = Tot++;
val[root] = 0;
if(l != r)
{
int mid = (l+r)>>1;
lson[root] = build(l,mid);
rson[root] = build(mid+1,r);
}
return root;
}
//如果那里发生改变则兴建一个节点而非像平常修改那个节点的值
int update(int root,int pos,int va)
{
int newroot = Tot++;
int tmp = newroot;
val[newroot] = val[root] + va;
int l = 1,r = m;
while(l < r)
{
int mid = (l+r)>>1;
if(pos <= mid)
{
lson[newroot] = Tot++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
lson[newroot] = lson[root];
rson[newroot] = Tot++;
newroot = rson[newroot];
root = rson[root];
l = mid+1;
}
val[newroot] = val[root] + va;
}
return tmp;
}
int query(int lt,int rt,int lca,int k)
{
int lca_rt = T[lca];
int pos = Hash(a[lca]);
int l = 1, r = m;
while(l < r)
{
int mid = (l+r)>>1;
int tmp = val[lson[lt]]+val[lson[rt]]-2*val[lson[lca_rt]]+(pos>=l&&pos<=mid);
if(tmp >= k)
{
lt = lson[lt];
rt = lson[rt];
lca_rt = lson[lca_rt];
r = mid;
}
else
{
k -= tmp;
l = mid+1;
lt = rson[lt];
rt = rson[rt];
lca_rt = rson[lca_rt];
}
}
return l;
}
int rmq[maxn*2]; //表示深度
struct ST
{
int mm[maxn*2];
int dp[maxn*2][20];
void ini(int n)
{
mm[0] = -1;
for(int i = 1; i <= n; i++)
{
mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];
dp[i][0] = i;
}
for(int j = 1; j <= mm[n]; j++)
for(int i = 1; i + (1<<j) - 1 <= n; i++)
dp[i][j] = rmq[dp[i][j-1]] < rmq[dp[i+(1<<(j-1))][j-1]]?
dp[i][j-1]:dp[i+(1<<(j-1))][j-1];
}
int query(int a,int b)
{
if(a > b)swap(a,b);
int k = mm[b-a+1];
return rmq[dp[a][k]] <= rmq[dp[b-(1<<k)+1][k]]?
dp[a][k]:dp[b-(1<<k)+1][k];
}
};
struct E
{
int to,next;
} edge[maxn*2];
int tot,head[maxn];
int F[maxn*2];
int P[maxn];
int cnt;
//F表示dfs的序列
//P[i]表示i第一次出现的位置
ST st;
void init() //初始化
{
Tot = tot = 0;
memset(head,-1,sizeof(head));
}
void dfs(int u,int pre,int dep)
{
F[++cnt] = u;
rmq[cnt] = dep;
P[u] = cnt;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(v == pre)continue;
dfs(v,u,dep+1);
F[++cnt] = u;
rmq[cnt] = dep;
}
}
void ini_lca(int root,int num)
{
cnt = 0;
dfs(root,root,0);
st.ini(2*num-1);
}
void addedge(int u,int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
int query_lca(int u,int v)
{
return F[st.query(P[u],P[v])];
}
void dfs_build(int u,int pre)
{
int pos = Hash(a[u]);
T[u] = update(T[pre],pos,1);
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(v == pre) continue;
dfs_build(v,u);
}
}
int main()
{
int q;
while(scanf("%d%d",&n,&q) == 2)
{
for(int i = 1; i <= n; i++)
scanf("%d",&a[i]);
ini_hash();
init();
int u,v,k;
for(int i = 1; i < n; i++)
{
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
}
ini_lca(1,n);
T[n+1] = build(1,m);
dfs_build(1,n+1);
while(q--)
{
scanf("%d%d%d",&u,&v,&k);
printf("%d
",t[query(T[u],T[v],query_lca(u,v),k)]);
}
}
return 0;
}