hdu 4348 主席树的区间更新
Background
To The Moon is a independent game released in November 2011, it is a role-playing adventure game powered by RPG Maker.
The premise of To The Moon is based around a technology that allows us to permanently reconstruct the memory on dying man. In this problem, we’ll give you a chance, to implement the logic behind the scene.
You‘ve been given N integers A 11, A 22,…, A NN. On these integers, you need to implement the following operations:
1. C l r d: Adding a constant d for every {A i | l <= i <= r}, and increase the time stamp by 1, this is the only operation that will cause the time stamp increase.
2. Q l r: Querying the current sum of {A i | l <= i <= r}.
3. H l r t: Querying a history sum of {A i | l <= i <= r} in time t.
4. B t: Back to time t. And once you decide return to a past, you can never be access to a forward edition anymore.
.. N, M ≤ 10 5, |A ii| ≤ 10 9, 1 ≤ l ≤ r ≤ N, |d| ≤ 10 4 .. the system start from time 0, and the first modification is in time 1, t ≥ 0, and won’t introduce you to a future state.
Input
n m
A 1 A 2 … A n
… (here following the m operations. )
Output
… (for each query, simply print the result. )
Sample Input
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
2 4
0 0
C 1 1 1
C 2 2 -1
Q 1 2
H 1 2 1
Sample Output
4
55
9
15
0
1
题意:
一个长度为n的数组,4种操作 :
(1)C l r d:区间[l,r]中的数都加1,同时当前的时间戳加1 。
(2)Q l r:查询当前时间戳区间[l,r]中所有数的和 。
(3)H l r t:查询时间戳t区间[l,r]的和 。
(4)B t:将当前时间戳置为t 。
所有操作均合法 。
很明显可以想到每个时间点是一颗线段树,由于有时间上的限制,所以用主席树,对于每棵线段树进行一般的区间操作即可,但是这里不能使用pushdown 因为节点很多(因为你更新到下一棵树的时候复制是上一棵树的如果pushdown要对很多的树进行pushdown,(因为查询的时间不一定),会很麻烦),所以是人工的更新lazy标记统计和,让其的值确定下来,成为一个静态的,对后面的树不再有影响。所以区间更新和询问的时候不能像一般线段树一样,而是要指定特定的区间,而且都是更新在不同是时间点上 例如 laz[now] 和 ls[now],rs[now]。查询的时候人工求和的过程是 找到指定的区间 每一层累加上 要找的区间*laz[now],还有不同的地方就是在pushup的地方要把(r-l+1)*laz 也统计上去sum除,这样查询到的最后的sum才是对的
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+100;
typedef long long ll;
ll a[N];
int ls[N*40],rs[N*40];
typedef long long ll;
ll sum[N*40],laz[N*40];
int root[N*40];
int tot;
void pushup(int l,int r,int now)
{
sum[now]=sum[ls[now]]+sum[rs[now]]+1LL*(r-l+1)*laz[now];
}
int build(int l,int r)
{
int now=++tot;
sum[now]=0;
laz[now]=0;
if(l==r)
{
sum[now]=a[l];
return now;
}
int mid=(l+r)>>1;
ls[now]=build(l,mid);
rs[now]=build(mid+1,r);
pushup(l,r,now);
return now;
}
void update(int pre,int &now,int L,int R,int l,int r,ll d)
{
now=++tot;
ls[now]=ls[pre];
rs[now]=rs[pre];
sum[now]=sum[pre];
laz[now]=laz[pre];
if(l==L&&R==r)
{
sum[now]+=1LL*(R-L+1)*d;
laz[now]+=d;
return ;
}
int mid=(l+r)>>1;
if(R<=mid) update(ls[pre],ls[now],L,R,l,mid,d);
else if(L>mid) update(rs[pre],rs[now],L,R,mid+1,r,d);
else{
update(ls[pre],ls[now],L,mid,l,mid,d);
update(rs[pre],rs[now],mid+1,R,mid+1,r,d);
}
pushup(l,r,now);
}
ll query(int now,int L,int R,int l,int r)
{
if(L==l&&R==r)
{
return sum[now];
}
ll res=1LL*(R-L+1)*laz[now];
int mid=(l+r)>>1;
if(R<=mid) res+=query(ls[now],L,R,l,mid);
else if(L>mid) res+=query(rs[now],L,R,mid+1,r);
else {
res+=query(ls[now],L,mid,l,mid);
res+=query(rs[now],mid+1,R,mid+1,r);
}
return res;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
tot=0;
int now=0;
for(int i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
}
root[0]=build(1,n);
for(int i=1;i<=m;i++){
char ch;
scanf(" %c",&ch);
if(ch=='C') {
int l,r;
ll d;
scanf("%d%d%lld",&l,&r,&d);
update(root[now],root[now+1],l,r,1,n,d);
now++;
}
else if(ch=='Q')
{
int l,r;
scanf("%d%d",&l,&r);
printf("%lld\n",query(root[now],l,r,1,n));
}
else if(ch=='H')
{
int l,r,d;
scanf("%d%d%d",&l,&r,&d);
printf("%lld\n",query(root[d],l,r,1,n));
}
else{
int d;
scanf("%d",&d);
now=d;
}
}
}
}