卧薪尝胆,厚积薄发。
TJOI2016&HEOI2016 排序
Date: Tue Aug 21 10:58:53 CST 2018 In Category: NoCategory

Description:

给出一个 $n$ 的全排列, $m$ 次操作将一个区间升序或降序排列,最后询问一次第 $p$ 个位置上的数是多少。
$1\le n \le 30000$

Solution:

先二分这个值,然后把序列中 $\ge$ 他的数赋成一,其他数赋成零,排序操作就是把这个区间中的一全部放在开始或结束,可以用线段树维护区间和和区间赋值。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int n,m;
#define MAXN 300100
int a[MAXN];
int pos;
struct opt
{
int type,l,r;
}p[MAXN];
int s[MAXN];
struct node
{
int lc,rc;
int sum,tag;
node(){tag = -1;sum = 0;}
}t[MAXN << 1];
int ptr = 0;
int newnode(){++ptr;t[ptr].tag = -1;return ptr;}
int root;
#define mid ((l + r) >> 1)
void build(int &rt,int l,int r)
{
rt = newnode();
if(l == r)
{
t[rt].sum = s[l];
return;
}
build(t[rt].lc,l,mid);
build(t[rt].rc,mid + 1,r);
t[rt].sum = t[t[rt].lc].sum + t[t[rt].rc].sum;
return;
}
void pushdown(int rt,int l,int r)
{
if(t[rt].tag == -1)return;
t[t[rt].lc].tag = t[t[rt].rc].tag = t[rt].tag;
t[t[rt].lc].sum = (mid - l + 1) * t[rt].tag;
t[t[rt].rc].sum = (r - mid) * t[rt].tag;
t[rt].tag = -1;
return;
}
void set(int rt,int L,int R,int k,int l,int r)
{
if(L > R)return;
if(L <= l && r <= R)
{
t[rt].sum = k * (r - l + 1);
t[rt].tag = k;
return;
}
pushdown(rt,l,r);
if(L <= mid)set(t[rt].lc,L,R,k,l,mid);
if(R > mid)set(t[rt].rc,L,R,k,mid + 1,r);
t[rt].sum = t[t[rt].lc].sum + t[t[rt].rc].sum;
return;
}
int query(int rt,int L,int R,int l,int r)
{
if(L <= l && r <= R)return t[rt].sum;
pushdown(rt,l,r);
int res = 0;
if(L <= mid)res += query(t[rt].lc,L,R,l,mid);
if(R > mid)res += query(t[rt].rc,L,R,mid + 1,r);
return res;
}
void pt(int rt,int l,int r)
{
if(l == r){cout << t[rt].sum << " ";return;}
pushdown(rt,l,r);
pt(t[rt].lc,l,mid);
pt(t[rt].rc,mid + 1,r);
return;
}
bool check(int k)
{
for(int i = 1;i <= n;++i)
{
if(a[i] >= k)s[i] = 1;
else s[i] = 0;
}
ptr = 0;
build(root,1,n);
for(int i = 1;i <= m;++i)
{
int sum1 = query(root,p[i].l,p[i].r,1,n),sum0 = p[i].r - p[i].l + 1 - sum1;
if(p[i].type == 0)
{
set(root,p[i].l,p[i].l + sum0 - 1,0,1,n);
set(root,p[i].l + sum0,p[i].r,1,1,n);
}
else
{
set(root,p[i].l,p[i].l + sum1 - 1,1,1,n);
set(root,p[i].l + sum1,p[i].r,0,1,n);
}
}
return (query(root,pos,pos,1,n) == 1);
}
int main()
{
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;++i)scanf("%d",&a[i]);
for(int i = 1;i <= m;++i)
{
scanf("%d%d%d",&p[i].type,&p[i].l,&p[i].r);
}
scanf("%d",&pos);
int l = 1,r = n,middle;
while(l < r)
{
middle = ((l + r + 1) >> 1);
if(check(middle))l = middle;
else r = middle - 1;
}
cout << l << endl;
return 0;
}
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