HDU 4348.To the moon SPOJ

To the moon

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 8372    Accepted Submission(s): 1986

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

 

 

 

题意：

长度为n的整数序列，支持四个操作

1：Q l r  输出区间[l,r]的总和

2：C l r x 区间[l,r]的每个值都增加x，此时时间增加1

3：H l r t 询问在t时刻区间[l,r]的总和

4：B t 时间回到t

延时标记如果下传的话，内存不够，所以要节省内存，按照正常的区间更新的操作，标记下传，就新开节点，这道题中，我们不新开节点，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，从根节点走到目标节点的过程中累加所经过节点上的标记值就可以。

所以就不能用以前的查询写法，if(L<=m) ... if(R> m) ... ，就需要换一种写法，就是这里:

　　 //if(L<=m) ret+=query(ls[rt],L,R,lson,c);
    //if(R> m) ret+=query(rs[rt],L,R,rson,c);
    if(R<=m) ret+=query(ls[rt],L,R,lson);
    else if(L> m) ret+=query(rs[rt],L,R,rson);
    else ret+=query(ls[rt],L,m,lson)+query(rs[rt],m+1,R,rson);
    return ret;

这道题真的自闭，wa了一晚上，最后发现，延时标记累加的时候，lazy[rt]*(R-L+1)写成了lazy[rt]*(r-l+1)。。。

其他的就是时间回到t的时候，直接将时间的值更新就可以。

代码:

//HDU 4348.To the moon-可持久化线段树-区间更新，延时标记不下传，优化空间
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define lson l,m
#define rson m+1,r

int rt[maxn],ls[maxn<<5],rs[maxn<<5],sz;
ll sum[maxn<<5],lazy[maxn<<5];

void pushup(int rt,int m)
{
    sum[rt]=sum[ls[rt]]+sum[rs[rt]]+(ll)lazy[rt]*m;
}

void build(int &rt,int l,int r)
{
    rt=++sz;lazy[rt]=0;
    if(l==r){
        scanf("%lld",&sum[rt]);
        return ;
    }

    int m=(l+r)>>1;
    build(ls[rt],lson);
    build(rs[rt],rson);
    pushup(rt,r-l+1);
}

void update(int pre,int &rt,int L,int R,int l,int r,ll c)
{
    rt=++sz;lazy[rt]=lazy[pre];sum[rt]=sum[pre];
    ls[rt]=ls[pre];rs[rt]=rs[pre];
    if(L<=l&&r<=R){
        lazy[rt]+=c;
        sum[rt]+=(ll)c*(r-l+1);
        return ;
    }

    int m=(l+r)>>1;
    if(L<=m) update(ls[pre],ls[rt],L,R,lson,c);
    if(R> m) update(rs[pre],rs[rt],L,R,rson,c);
    pushup(rt,r-l+1);
}

ll query(int rt,int L,int R,int l,int r)
{
    if(L<=l&&r<=R){
        return sum[rt];
    }

    //ll ret=(ll)lazy[rt]*(r-l+1);
    ll ret=(ll)lazy[rt]*(R-L+1);
    int m=(l+r)>>1;
    //if(L<=m) ret+=query(ls[rt],L,R,lson,c);
    //if(R> m) ret+=query(rs[rt],L,R,rson,c);
    if(R<=m) ret+=query(ls[rt],L,R,lson);
    else if(L> m) ret+=query(rs[rt],L,R,rson);
    else ret+=query(ls[rt],L,m,lson)+query(rs[rt],m+1,R,rson);
    return ret;
}

char op[5];

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        sz=0;memset(rt,0,sizeof(rt));
        build(rt[0],1,n);
        int tag=0;
        for(int i=1;i<=m;i++){
            scanf("%s",op);
            if(op[0]=='C'){
                int l,r;ll c;
                scanf("%d%d%lld",&l,&r,&c);
                tag++;
                update(rt[tag-1],rt[tag],l,r,1,n,c);
            }
            else if(op[0]=='Q'){
                int l,r;
                scanf("%d%d",&l,&r);
                printf("%lld
",query(rt[tag],l,r,1,n));
            }
            else if(op[0]=='H'){
                int l,r,d;
                scanf("%d%d%d",&l,&r,&d);
                printf("%lld
",query(rt[d],l,r,1,n));
            }
            else if(op[0]=='B'){
                int x;
                scanf("%d",&x);
                tag=x;
            }
        }
    }
    return 0;
}

讲道理，这题还是没完全弄明白，还是有点迷糊。

mdzz。。。