# BZOJ 2588 Count on a tree (COT) 是持久的段树

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 100010
#define NIL (tree[0])
using namespace std;

pair<int,int> src[MAX];

struct Complex{
Complex *son[2];
int cnt;

Complex(Complex *_,Complex *__,int ___):cnt(___) {
son[0] = _;
son[1] = __;
}
Complex() {}
}*tree[MAX];

int xx[MAX];

int next[MAX << 1],aim[MAX << 1];

int father[MAX][20],deep[MAX];

void DFS(int x);
void SparseTable();

Complex *BuildTree(Complex *pos,int x,int y,int val);
int GetAns(Complex *l,Complex *r,Complex *f,Complex *p,int x,int y,int k);
int GetLCA(int x,int y);

int main()
{
for(int i = 1;i <= points; ++i)
scanf("%d",&src[i].first),src[i].second = i;
sort(src + 1,src + points + 1);
for(int i = 1;i <= points; ++i)
xx[src[i].second] = i;
for(int x,y,i = 1;i < points; ++i) {
scanf("%d%d",&x,&y);
}
NIL = new Complex(NULL,NULL,0);
NIL->son[0] = NIL->son[1] = NIL;
DFS(1);
SparseTable();
int now = 0;
for(int x,y,k,i = 1;i <= asks; ++i) {
scanf("%d%d%d",&x,&y,&k);
}
return 0;
}

{
aim[total] = y;
}

void DFS(int x)
{
deep[x] = deep[father[x][0]] + 1;
tree[x] = BuildTree(tree[father[x][0]],1,points,xx[x]);
for(int i = head[x];i;i = next[i]) {
if(aim[i] == father[x][0])	continue;
father[aim[i]][0] = x;
DFS(aim[i]);
}
}

void SparseTable()
{
for(int j = 1;j <= 19; ++j)
for(int i = 1;i <= points; ++i)
father[i][j] = father[father[i][j - 1]][j - 1];
}

{
int lca = GetLCA(x,y);
return GetAns(tree[x],tree[y],tree[lca],tree[father[lca][0]],1,points,k);
}

Complex *BuildTree(Complex *pos,int x,int y,int val)
{
int mid = (x + y) >> 1;
if(x == y)	return new Complex(NIL,NIL,pos->cnt + 1);
if(val <= mid)	return new Complex(BuildTree(pos->son[0],x,mid,val),pos->son[1],pos->cnt + 1);
return new Complex(pos->son[0],BuildTree(pos->son[1],mid + 1,y,val),pos->cnt + 1);
}

int GetLCA(int x,int y)
{
if(deep[x] < deep[y])	swap(x,y);
for(int i = 19; ~i; --i)
if(deep[father[x][i]] >= deep[y])
x = father[x][i];
if(x == y)	return x;
for(int i = 19; ~i; --i)
if(father[x][i] != father[y][i])
x = father[x][i],y = father[y][i];
return father[x][0];
}

int GetAns(Complex *l,Complex *r,Complex *f,Complex *p,int x,int y,int k)
{
if(x == y)	return x;
int mid = (x + y) >> 1;
int temp = l->son[0]->cnt + r->son[0]->cnt - f->son[0]->cnt - p->son[0]->cnt;
if(k <= temp)	return GetAns(l->son[0],r->son[0],f->son[0],p->son[0],x,mid,k);
return GetAns(l->son[1],r->son[1],f->son[1],p->son[1],mid + 1,y,k - temp);
}

posted @ 2015-07-23 19:26  yxwkaifa  阅读(146)  评论(0编辑  收藏  举报