hdu 4348 To the moon(主席树,区间更新节省内存,经典)

题目链接

To the moon

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5038    Accepted Submission(s): 1134

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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 42 40 0C 1 1 1C 2 2 -1Q 1 2H 1 2 1

 

Sample Output

45591501

 

Author

HIT

 

Source

2012 Multi-University Training Contest 5 

参考题解

这一题和这个题是一样的，唯一区别就是内存限制变小了，当然还有输出时多组数据之间不用空行。

于是，学到了一种函数式线段树成段更新时节约内存的办法 。。。

先考虑朴素的仅支持成段加减的线段树，我们可以用方式解决：

1.正常的懒惰标记，当访问到带有懒惰标记节点的子区间时将标记下传；

2.不用下传的懒惰标记，我们用一个标记来记录当前节点的整段区间被累加了多少，当询问的时候我们在从根节点走到目标结点的过程中不断累加所经过节点上的标记值。。。

基于这两种思想，就有了函数式线段树的两种实现方式，第一种在pushdown的过程中会产生大量的结点，而第二种没有pushdown的过程，不会新建过多的节点

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>#include<queue>#include<stack>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const int maxn=1e5+10;struct node{    int l,r;    ll sum,add;}T[maxn*30];int root[maxn];int cnt;void pushup(int x){    T[x].sum=T[T[x].l].sum+T[T[x].r].sum;}int build(int l,int r){    int root=++cnt;    T[root].add=0;    if(l==r)    {        scanf("%lld",&T[root].sum);        T[root].l=T[root].r=0;        return root;    }    int m=(l+r)/2;    T[root].l=build(l,m),T[root].r=build(m+1,r);    pushup(root);    return root;}void copy(int x,int y){    T[x].l=T[y].l;    T[x].r=T[y].r;    T[x].sum=T[y].sum;    T[x].add=T[y].add;}int update(int root,int L,int R,int l,int r,int v){    int now=++cnt;    copy(now,root);    T[now].sum+=1ll*v*(r-l+1);    if(L==l&&R==r)    {        T[now].add+=v;        return now;    }    int m=(L+R)/2;    if(r<=m) T[now].l=update(T[root].l,L,m,l,r,v);    else if(l>m) T[now].r=update(T[root].r,m+1,R,l,r,v);    else    {        T[now].l=update(T[root].l,L,m,l,m,v);        T[now].r=update(T[root].r,m+1,R,m+1,r,v);    }    return now;}ll query(int root,int L,int R,int l,int r){    ll ans=0;    if(L==l&&r==R)        return T[root].sum;    int m=(L+R)/2;    ans+=(r-l+1)*T[root].add;    if(r<=m) return ans+query(T[root].l,L,m,l,r);    else if(l>m) return ans+query(T[root].r,m+1,R,l,r);    else return ans+query(T[root].l,L,m,l,m)+query(T[root].r,m+1,R,m+1,r);    return ans;}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        int now=0;        cnt=0;        root[now]=build(1,n);        char op[5];        while(m--)        {            scanf("%s",op);            int l,r,v;            if(op[0]=='Q')            {                scanf("%d%d",&l,&r);                printf("%lld\n",query(root[now],1,n,l,r));            }            else if(op[0]=='C')            {                scanf("%d%d%d",&l,&r,&v);                now++;                root[now]=update(root[now-1],1,n,l,r,v);            }            else if(op[0]=='H')            {                scanf("%d%d%d",&l,&r,&v);                printf("%lld\n",query(root[v],1,n,l,r));            }            else if(op[0]=='B')                scanf("%d",&v),now=v;        }        //puts("");    }    return 0;}