hdu 4348 To the moon 主席树区间更新

To the moon

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)


[align=left]Problem Description[/align]
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[1], A[2],..., A
. On these integers, you need to implement the following operations:
1. C l r d: Adding a constant d for every {Ai | 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 {Ai | l <= i <= r}.
3. H l r t: Querying a history sum of {Ai | 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 ≤ 105, |A[i]| ≤ 109, 1 ≤ l ≤ r ≤ N, |d| ≤ 104 .. 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.

[align=left]Input[/align]
n m
A1 A2 ... An
... (here following the m operations. )

[align=left]Output[/align]
... (for each query, simply print the result. )

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

4
55
9
15

0
1

[align=left]Author[/align]
HIT

[align=left]Source[/align]
2012 Multi-University Training Contest 5

给你一个数组，让你维护，m次操作
　　询问当前时刻一个区间的和
　　询问在t时刻的一个区间的和
　　回到t时刻
　　时间+1，在此时刻+1下更新一个区间的值

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=5e5+30010,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
ll a
;
struct Chairmantree
{
int rt[N*20],ls[N*20],rs[N*20];
ll sum[N*20],lazy[N*20];
int tot;
void init()
{
tot=0;
}
void pushup(int l,int r,int pos)
{
sum[pos]=sum[ls[pos]]+sum[rs[pos]]+1LL*(r-l+1)*lazy[pos];
}
void build(int l,int r,int &pos)
{
pos=++tot;
lazy[pos]=0;
sum[pos]=0;
if(l==r)
{
sum[pos]=a[l];
return;
}
int mid=(l+r)>>1;
build(l,mid,ls[pos]);
build(mid+1,r,rs[pos]);
pushup(l,r,pos);
}
void update(int L,int R,ll c,int pre,int l,int r,int &pos)
{
pos=++tot;
ls[pos]=ls[pre];
rs[pos]=rs[pre];
sum[pos] = sum[pre];
lazy[pos]=lazy[pre];
if(L==l&&r==R)
{
sum[pos]+=1LL*(r-l+1)*c;
lazy[pos]+=c;
return;
}
int mid=(l+r)>>1;
if(R<=mid)
update(L,R,c,ls[pre],l,mid,ls[pos]);
else if(L>mid)
update(L,R,c,rs[pre],mid+1,r,rs[pos]);
else
{
update(L,mid,c,ls[pre],l,mid,ls[pos]);
update(mid+1,R,c,rs[pre],mid+1,r,rs[pos]);
}
pushup(l,r,pos);
}
ll query(int L,int R,int l,int r,int pos)
{
if(L==l&&r==R)
return sum[pos];
int mid=(l+r)>>1;
ll ans=1LL*lazy[pos]*(R-L+1);
if(R<=mid)
ans+=query(L,R,l,mid,ls[pos]);
else if(L>mid)
ans+=query(L,R,mid+1,r,rs[pos]);
else
{
ans+=query(L,mid,l,mid,ls[pos]);
ans+=query(mid+1,R,mid+1,r,rs[pos]);
}
return ans;
}
};
Chairmantree tree;
char s[10];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=1; i<=n; i++)
scanf("%lld",&a[i]);
tree.init();
tree.build(1,n,tree.rt[0]);
int now=0;
while(m--)
{
scanf("%s",s);
if(s[0]=='C')
{
int l,r;
ll c;
scanf("%d%d%lld",&l,&r,&c);
tree.update(l,r,c,tree.rt[now],1,n,tree.rt[now+1]);
now++;
}
else if(s[0]=='Q')
{
int l,r;
scanf("%d%d",&l,&r);
printf("%lld\n",tree.query(l,r,1,n,tree.rt[now]));
}
else if(s[0]=='H')
{
int l,r,t;
scanf("%d%d%d",&l,&r,&t);
printf("%lld\n",tree.query(l,r,1,n,tree.rt[t]));
}
else
{
int x;
scanf("%d",&x);
now=x;
}
}
}
return 0;
}