hdu 5193 Go to movies Ⅱ 块状链表 ＋ 树状数组

数组的定位o(1),插入o(n). 链表的定位o(n),插入o(1).

所以把二者结合，是复杂度均摊为 sqrt(n)

设每块的大小为S，那么删除或者添加元素时，维护逆序对数的复杂度是O(S+\frac{P}{S}*
\log n)o（S+(p/s)*logn），S是块内直接暴力更新逆序对的代价，(n/s)∗logn在前面块找比它大和在后面块中找比它小的代价,P表示当前元素的个数。为了使这两部分复杂度尽量均摊让S=\frac{P}{S}*
\log nS=(p/s)*logn，S取sqrt(p*logn)。直接通过分块暴力添加和删除时，块的大小会退化，。。。。。（注：原官方题解说重构，不太清楚怎么叫重构。这里为了防止退化，在每个块元素过多时采取分裂操作）。因此整个问题的复杂度为O(m\sqrt{n\log
n})O(m*sqrt(n*logn)).

/*************************************************************************
> File Name: a.cpp
> Author: TechMonster
> Mail: 928221136@qq.com
> Created Time: 二  7/ 5 09:14:50 2016
************************************************************************/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<string>
#include<math.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
#define ls (o<<1)
#define rs (o<<1|1)
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x, y) memcpy(x, y, sizeof(x))
#define MP(x,y) make_pair(x,y)
#define PB(x) push_back(x)
#define FF first
#define SS second
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 50010;
const int lim = 150;

void add(int *c,int x,int v)
{
while(x <= 20000)
{
c[x] += v;
x += x&-x;
}
}
int sum(int *c,int x)
{
int ret = 0;
while(x>0)
{
ret += c[x];
x -= x&-x;
}
return ret;
}
struct data
{
int s,a[305],c[20010];
data *nxt;
data()
{
MS(c,0);
nxt = NULL;
}
};
int n,m;
data *root;
void insert(int x,int pos)
{
data *now = root;
if(root == NULL)
{
root = new data;
root->s = 1;
root->a[1] = x;
add(root->c,x,1);
return;
}
while(pos > now->s && now->nxt != NULL)
{
pos -= now->s;
now = now->nxt;
}
memmove(now->a+pos+1,now->a+pos,sizeof(int)*(now->s-pos+1));
now->a[pos] = x;
add(now->c,x,1);
now->s++;
if(now->s == 2*lim)//分裂
{
data *lst = new data;
lst->nxt = now->nxt;
now->nxt = lst;
memcpy(lst->a+1,now->a+lim+1,sizeof(int)*lim);
lst->s = now->s = lim;
for(int i = 1; i <= lim; ++i)
{
add(now->c,lst->a[i],-1);
add(lst->c,lst->a[i],1);
}
}
}
int find(int pos)
{
data *now = root;
while(pos > now->s && now->nxt != NULL)
{
pos -= now->s;
now = now->nxt;
}
return now->a[pos];
}
int calc(int pos)
{
data *now = root;
int ret = 0;
int x = find(pos);
while(pos > now->s && now->nxt != NULL)
{
ret += sum(now->c,20000) - sum(now->c,x);
pos -= now->s;
now = now->nxt;
}
for(int i = 1; i < pos; ++i)
if(now->a[i] > x) ret++;
for(int i = pos+1; i <= now->s; ++i)
if(now->a[i] < x) ret++;
while(now->nxt != NULL)
{
now = now->nxt;
ret += sum(now->c,x-1);
}
return ret;
}
void del(int pos)
{
data *now = root;
while(pos > now->s && now->nxt != NULL)
{
pos -= now->s;
now = now->nxt;
}
add(now->c,now->a[pos],-1);
memmove(now->a+pos,now->a+pos+1,sizeof(int)*(now->s-pos));
now->s--;
}
void destroy(data *now)
{
if(now->nxt != NULL) destroy(now->nxt);
delete now;
}
void solve()
{
root = NULL;//delete 后 ，root会指向非法内存，刚开始忘了初始化一直re
int ope,u,v,ans = 0;
int x,y;
for(int i = 1;  i <= n; ++i)
{
scanf("%d",&u);
insert(u,i);
ans += calc(i);
}
for(int i = 1; i <= m; ++i)
{
scanf("%d",&ope);
if(ope == 1)
{
scanf("%d",&x);
ans -= calc(x);
del(x);
}
else
{
scanf("%d%d",&x,&y);
insert(y,x+1);
ans += calc(x+1);
}
printf("%d\n",ans);
}
destroy(root);
}
int main()
{
while(~scanf("%d%d",&n,&m))
solve();
return 0;
}