hdu4348 To the moon

地址：http://acm.hdu.edu.cn/showproblem.php?pid=4348

题目：

To the moon

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5374    Accepted Submission(s): 1232

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

 

　思路：

　　持久化线段树：每次根据前一个版本建树，遇到要修改的区间就新增加节点，不修改的沿用前一版本的节点。

　　怎么区间修改呢？

　　　　打标记即可，但标记不要下传！否则空间复杂度会炸。

　　那怎么办？

　　　　在修改时经过的所有区间需要维护修改后的区间和，修改标记打在被整段覆盖的区间上。

　　标记不下传，那查询怎么办？

　　　　查询到一个区间时，把之前的遇到的区间的标记求和，答案就是标记和乘以区间长度+这个区间的区间和。

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=5e6+7;
const int mod=1e9+7;

int tot,ls[K],rs[K],rt[K];
LL sum[K],add[K],v[K];
//sum[o]记录的是该节点区间内出现的数的个数
//区间指的是将数离散化后的区间
void build(int &o,int l,int r)
{
    o=tot++,sum[o]=v[l],add[o]=0;
    if(l==r) return ;
    int mid=l+r>>1;
    build(ls[o],l,mid),build(rs[o],mid+1,r);
    sum[o]=sum[ls[o]]+sum[rs[o]];
}
void update(int &o,int p,int l,int r,int nl,int nr,LL x)
{
    o=tot++,ls[o]=ls[p],rs[o]=rs[p],sum[o]=sum[p]+(nr-nl+1)*x,add[o]=add[p];
    if(l==nl && r==nr){add[o]+=x;return;}
    int mid=l+r>>1;
    if(nr<=mid) update(ls[o],ls[p],l,mid,nl,nr,x);
    else if(nl>mid) update(rs[o],rs[p],mid+1,r,nl,nr,x);
    else update(ls[o],ls[p],l,mid,nl,mid,x),update(rs[o],rs[p],mid+1,r,mid+1,nr,x);
    sum[o]=sum[ls[o]]+sum[rs[o]]+(r-l+1)*add[o];
}
LL query(int o,int l,int r,int nl,int nr,LL f)
{
    if(l==nl && r==nr) return sum[o]+(nr-nl+1)*f;
    int mid=l+r>>1;
    if(nr<=mid) return query(ls[o],l,mid,nl,nr,f+add[o]);
    else if(nl>mid) return query(rs[o],mid+1,r,nl,nr,f+add[o]);
    return query(ls[o],l,mid,nl,mid,f+add[o])+query(rs[o],mid+1,r,mid+1,nr,f+add[o]);
}
int main(void)
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        tot=0;
        for(int i=1;i<=n;i++)
            scanf("%I64d",v+i);
        build(rt[0],1,n);
        int k=0,l,r,d;
        char op[10];
        while(m--)
        {
            scanf("%s",op);
            if(op[0]=='C')
                scanf("%d%d%d",&l,&r,&d),update(rt[k+1],rt[k],1,n,l,r,d),k++;
            else if(op[0]=='Q')
                scanf("%d%d",&l,&r),printf("%I64d\n",query(rt[k],1,n,l,r,0));
            else if(op[0]=='H')
                scanf("%d%d%d",&l,&r,&d),printf("%I64d\n",query(rt[d],1,n,l,r,0));
            else
                scanf("%d",&k);
        }
    }
    return 0;
}