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)

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 ≤ 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.

 

Input

n m
A₁ A₂ ... 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

 

Author

HIT

 

Source

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[N];
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
",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
",tree.query(l,r,1,n,tree.rt[t]));
            }
            else
            {
                int x;
                scanf("%d",&x);
                now=x;
            }
        }
    }
    return 0;
}