bzoj2588: Spoj 10628. Count on a tree

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

题意：中文题。

分析：树上建可持久化线段树，从父亲那继承。查询的时候减去lca和lca的父亲两个前缀即可。详见代码。O(nlogn+mlogn) PS：注意输出格式！

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=100100;
const int MAX=100000000;
const int mod=100000000;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=998244353;
const ll INF=10000000010;
typedef double db;
typedef unsigned long long ull;
int a
,b
,c
;
int tot,u
,v[2*N],pre[2*N];
void add(int x,int y) {
v[tot]=y;pre[tot]=u[x];u[x]=tot++;
}
int n,de
,fa
[20];
int sz,root
,ls[20*N],rs[20*N],sum[20*N];
void build(int l,int r,int x,int &y,int z) {
y=++sz;sum[y]=sum[x]+1;
if (l==r) return ;
int mid=(l+r)>>1;
ls[y]=ls[x];rs[y]=rs[x];
if (z<=mid) build(l,mid,ls[x],ls[y],z);
else build(mid+1,r,rs[x],rs[y],z);
}
void dfs(int x,int y) {
de[x]=de[y]+1;fa[x][0]=y;
build(1,n,root[y],root[x],a[x]);
for (int i=1;i<=20;i++)
if (de[x]>1<<i) fa[x][i]=fa[fa[x][i-1]][i-1];
for (int i=u[x];i!=-1;i=pre[i])
if (v[i]!=y) dfs(v[i],x);
}
int getlca(int x,int y) {
if (de[x]<de[y]) { x^=y;y^=x;x^=y; }
for (int i=20;i>=0;i--)
if (de[x]-(1<<i)>=de[y]) x=fa[x][i];
if (x==y) return x;
for (int i=20;i>=0;i--)
if (de[x]>1<<i&&fa[x][i]!=fa[y][i]) { x=fa[x][i];y=fa[y][i]; }
return fa[x][0];
}
int get_k(int l,int r,int x,int y,int z,int w,int k) {
if (l==r) return l;
int mid=(l+r)>>1;
if (sum[ls[z]]+sum[ls[w]]-sum[ls[x]]-sum[ls[y]]>=k) return get_k(l,mid,ls[x],ls[y],ls[z],ls[w],k);
else return get_k(mid+1,r,rs[x],rs[y],rs[z],rs[w],k-(sum[ls[z]]+sum[ls[w]]-sum[ls[x]]-sum[ls[y]]));
}
int main()
{
int i,m,k,x,y,w,lca,ans=0;
scanf("%d%d", &n, &m);
for (i=1;i<=n;i++) {
scanf("%d", &a[i]);b[i]=a[i];
}
sort(b+1,b+n+1);
k=unique(b+1,b+n+1)-(b+1);
for (i=1;i<=n;i++) {
w=lower_bound(b+1,b+k+1,a[i])-b;
c[w]=a[i];a[i]=w;
}
tot=0;memset(u,-1,sizeof(u));
for (i=1;i<n;i++) {
scanf("%d%d", &x, &y);
add(x,y);add(y,x);
}
dfs(1,0);
while (m--) {
scanf("%d%d%d", &x, &y, &k);
x^=ans;lca=getlca(x,y);
ans=c[get_k(1,n,root[lca],root[fa[lca][0]],root[x],root[y],k)];
printf("%d", ans);
if (m) printf("\n");
}
return 0;
}