HDU 4348 To the moon（可持久化线段树）

To the moon

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4287    Accepted Submission(s): 923

Problem Description

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 _[N]. 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 _[i]| ≤ 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

可持久化线段树，这道题目会卡内存，所以在pushdown和pushup的时候，新建节点可能会超内存

那么我就可以跳过pushup和pushdown

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
typedef long long int LL;
const int maxn=1e5;
int rt[maxn*35+5];
int ls[maxn*35+5];
int rs[maxn*35+5];
int p;
LL sum[maxn*35+5];
LL pos[maxn*35+5];
int n,m,t;
int newnode()
{
    ls[p]=rs[p]=sum[p]=pos[p]=0;
    return p++;
}
void build(int &node,int begin,int end)
{
    if(!node) node=newnode();
    if(begin==end)
    {
        scanf("%lld",&sum[node]);
        return;
    }
    int mid=(begin+end)>>1;
    build(ls[node],begin,mid);
    build(rs[node],mid+1,end);
    sum[node]=sum[ls[node]]+sum[rs[node]];
}
void update(int &node,int begin,int end,int left,int right,int val)
{
    sum[p]=sum[node];ls[p]=ls[node];rs[p]=rs[node];
    pos[p]=pos[node];
    node=p;p++;
    sum[node]+=1LL*val*(right-left+1);
    if(left==begin&&end==right)
    {
        pos[node]+=val;
        return;
    }
    int mid=(begin+end)>>1;
    if(right<=mid) update(ls[node],begin,mid,left,right,val);
    else if(left>mid) update(rs[node],mid+1,end,left,right,val);
    else
    {
        update(ls[node],begin,mid,left,mid,val);
        update(rs[node],mid+1,end,mid+1,right,val);
    }
}
LL query(int node,int begin,int end,int left,int right)
{
    if(left<=begin&&end<=right) return sum[node];
    LL ret=1LL*pos[node]*(right-left+1);
    int mid=(begin+end)>>1;
    if(right<=mid)  ret+=query(ls[node],begin,mid,left,right);
    else if(left>mid)  ret+=query(rs[node],mid+1,end,left,right);
    else
    {
        ret+=(query(ls[node],begin,mid,left,mid)+query(rs[node],mid+1,end,mid+1,right));

    }
    return ret;
}
int main()
{
    char x;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        t=0;
        p=0;
    build(rt[0],1,n);
    int l,r,d,time;
    
    LL ans;
    for(int i=1;i<=m;i++)
    {
        cin>>x;
        if(x=='C')
        {
            scanf("%d%d%d",&l,&r,&d);
            update(rt[++t]=rt[t-1],1,n,l,r,d);
        }
        else if(x=='Q')
        {
            scanf("%d%d",&l,&r);
            ans=query(rt[t],1,n,l,r);
            printf("%lld\n",ans);
        }
        else if(x=='H')
        {
            scanf("%d%d%d",&l,&r,&time);
            ans=query(rt[time],1,n,l,r);
            printf("%lld\n",ans);
        }
        else
        {
            scanf("%d",&time);
            t=time;
        }

    }
    }
    return 0;
}