POJ 2761 SBT(size balanced tree)

题意：给出n个数字m个询问，每次询问在范围[a,b]内第k小数字是多少，并输出。

思路：网上的最优解法是划分树或者归并树，最近学了SBT(size balanced tree)，正好里面有类似的select(t,k)选择第k小数的操作，所以就用SBT写了一下。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 1000005
#define inf 1<<28
#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define FOR(i,s,t) for(int i=(s);i<=(t);++i)
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
using namespace std;

struct SBT
{
int left , right , num ,size;
}tree[Max];

void left_rot(int &x)//左旋
{
int y = tree[x].right;
tree[x].right = tree[y].left;
tree[y].left = x;
tree[y].size = tree[x].size;
tree[x].size = tree[tree[x].left].size + tree[tree[x].right].size + 1;
x = y;
}

void right_rot(int &x)//右旋
{
int y = tree[x].left;
tree[x].left = tree[y].right;
tree[y].right = x;
tree[y].size = tree[x].size;
tree[x].size = tree[tree[x].left].size + tree[tree[x].right].size + 1;
x = y;
}

void maintain(int &x,bool flag)//更新
{
if(!flag)//左子树是0
{
if(tree[tree[tree[x].left].left].size > tree[tree[x].right].size)
right_rot(x);
else if(tree[tree[tree[x].left].right].size > tree[tree[x].right].size)
{
left_rot(tree[x].left);
right_rot(x);
}
else return ;
}
else//右子树是1
{
if(tree[tree[tree[x].right].right].size > tree[tree[x].left].size)
left_rot(x);
else if(tree[tree[tree[x].right].left].size > tree[tree[x].left].size)
{
right_rot(tree[x].right);
left_rot(x);
}
else return ;
}
maintain(tree[x].left,0);
maintain(tree[x].right,1);
maintain(x,1);
maintain(x,0);
}

int root , top;
void insert(int &x,int num)//插入
{
if(x == 0)
{
x = ++top;
tree[x].left = tree[x].right = 0;
tree[x].size = 1;
tree[x].num = num ;
}
else
{
tree[x].size ++;
if(num < tree[x].num )insert(tree[x].left , num);
else insert(tree[x].right , num);
maintain(x,num >= tree[x].num);
}
}

int select(int &x,int num)//选择第num个小数
{
int r = tree[tree[x].left].size + 1;
if(r == num )return tree[x].num;
else if (r > num)return select(tree[x].left,num);
else return select(tree[x].right,num - r);
}
int del(int &x,int num)//删除num值
{
int d_num ;
if(!x)return 0;
tree[x].size --;
if(num == tree[x].num ||(tree[x].num > num && tree[x].left == 0)||(tree[x].num < num && tree[x].right == 0))
{
d_num = tree[x].num;
if( tree[x].left && tree[x].right )
{
tree[x].num = del(tree[x].left,tree[x].num + 1);
}
else
{
x = tree[x].left + tree[x].right ;
}
}
else if(num > tree[x].num)
d_num = del(tree[x].right , num);
else if(num < tree[x].num )
d_num = del(tree[x].left ,num);
return d_num;
}
int num[100005];
struct kdq
{
int s,e,idx,ans,th;
}ans[50005];//代表询问，[s,e]是区间.th是第几,ans是输出的结果,idx是该询问的次序。
bool cmp(kdq x,kdq y)//按询问区间从前到后排
{
if(x.s == y.s)
return x.e<y.e;
return x.s<y.s;
}

bool cmp1(kdq x,kdq y)//按询问次序排，最后按顺序输出。
{
return x.idx<y.idx;
}

int main()
{
int n , m;
while(scanf("%d%d",&n,&m) != EOF)
{
for (int i = 1 ; i <= n ;i ++)scanf("%d",&num[i]);
for (int i = 0 ;i < m ;i ++){
scanf("%d%d%d",&ans[i].s,&ans[i].e,&ans[i].th);
ans[i].idx = i ;
}
sort(ans,ans + m ,cmp);
root = top =0;
for (int i = ans[0].s ; i <=ans[0].e ; i ++)insert(root , num[i]);
ans[0].ans = select(root , ans[0].th);
for (int i = 1; i < m ;i ++)
{
if(ans[i].s >= ans[i-1].e)//如果后面区间和前面的区间完全没有交集，则删除所有前面区间元素，插入所有后面区间的元素。
{
for (int j = ans[i-1].s ; j <= ans[i-1].e ;j ++)del(root , num[j]);
for (int j = ans[i].s ; j <= ans[i].e ; j ++)insert(root ,num[j]);
}
else if(ans[i].e <= ans[i-1].e)//如果后面的区间包含于前面的区间，则删除前面区间多余的元素
{
for (int j = ans[i-1].s ; j < ans[i].s ; j ++)del(root ,num[j]);
for (int j = ans[i].e + 1 ; j <=ans[i-1].e ;j ++ )del(root,num[j]);
}
else//如果后面的区间和前面的区间有部分交集，则删除前面区间多余的元素，插入后面区间多出的元素
{
for (int j = ans[i-1].s ; j < ans[i].s ; j ++)del(root ,num[j]);
for (int j = ans[i-1].e + 1; j <= ans[i].e ;j ++)insert(root ,num[j]);
}
//这里解释有点繁琐，但是不难想的
ans[i].ans = select(root ,ans[i].th);//记录结果
}
sort(ans,ans + m ,cmp1);
for (int i = 0 ;i < m ; i ++)printf("%d\n",ans[i].ans);
}
return 0;
}

同类型的题POJ 2104，用此方法TLE了。继续去学习其他树。。