[HDU 4348]To the moon

Posted 2020-10-08 Mafia

[HDU 4348]To the moon

题目

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

OUTPUT

4

55

9

15

 

0

1

解题报告

可。。。可。。。可持久化？

初识可持久化线段树+区间修改

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 using namespace std;
  5 typedef long long L;
  6 inline int read(){
  7     int sum(0),f(1);
  8     char ch(getchar());
  9     for(;ch<‘0‘||ch>‘9‘;ch=getchar())
 10         if(ch==‘-‘)
 11             f=-1;
 12     for(;ch>=‘0‘&&ch<=‘9‘;sum=sum*10+(ch^48),ch=getchar());
 13     return sum*f;
 14 }
 15 int n,m,t,cnt;
 16 int v[100005];
 17 int rt[100005],lch[4000005],rch[4000005];
 18 L sum[4000005],add[4000005];
 19 char op[2];
 20 inline int build(int l,int r){
 21     int x(++cnt);
 22     add[x]=0;
 23     if(l==r){
 24         sum[x]=v[l];
 25         lch[x]=rch[x]=0;
 26 //        cout<<"pos "<<x<<" "<<l<<‘ ‘<<r<<" "<<v[l]<<‘ ‘<<sum[x]<<endl;
 27         return x;
 28     }
 29     int mid((l+r)>>1);
 30     lch[x]=build(l,mid);
 31 //    cout<<"left child"<<x<<" "<<lch[x]<<‘ ‘<<sum[lch[x]]<<endl;
 32     rch[x]=build(mid+1,r);
 33 //    cout<<"right child"<<x<<" "<<rch[x]<<‘ ‘<<sum[rch[x]]<<endl;
 34 //    cout<<"wtf"<<endl;
 35     sum[x]=sum[lch[x]]+sum[rch[x]];
 36 //    cout<<"wtf"<<endl;
 37 //    cout<<x<<" "<<l<<‘ ‘<<r<<" "<<lch[x]<<‘ ‘<<rch[x]<<‘ ‘<<sum[lch[x]]<<‘ ‘<<sum[rch[x]]<<‘ ‘<<sum[x]<<endl;
 38     return x;
 39 }
 40 inline void pushup(int x,int len){
 41     sum[x]=sum[lch[x]]+sum[rch[x]]+add[lch[x]]*(len-(len>>1))+add[rch[x]]*(len>>1);
 42 }
 43 inline int update(int rt,int ll,int rr,L w,int l,int r){
 44     int newrt(++cnt);
 45     add[newrt]=add[rt];
 46     if(ll<=l&&r<=rr){
 47         sum[newrt]=sum[rt];
 48         add[newrt]=add[rt]+w;
 49         lch[newrt]=lch[rt];
 50         rch[newrt]=rch[rt];
 51         return newrt;
 52     }
 53     int mid((l+r)>>1);
 54     if(ll<=mid)
 55         lch[newrt]=update(lch[rt],ll,rr,w,l,mid);
 56     else
 57         lch[newrt]=lch[rt];
 58     if(mid<rr)
 59         rch[newrt]=update(rch[rt],ll,rr,w,mid+1,r);
 60     else
 61         rch[newrt]=rch[rt];
 62     pushup(newrt,r-l+1);
 63     return newrt;
 64 }
 65 inline L query(int rt,int ll,int rr,int l,int r,L add1){
 66 //    cout<<rt<<‘ ‘<<ll<<‘ ‘<<rr<<‘ ‘<<l<<‘ ‘<<r<<‘ ‘<<add1<<endl;
 67     if(ll<=l&&r<=rr)
 68         return sum[rt]+(add1+add[rt])*(r-l+1);
 69     int mid((l+r)>>1);
 70     L ret(0);
 71     if(ll<=mid)
 72         ret+=query(lch[rt],ll,rr,l,mid,add[rt]+add1);
 73     if(mid<rr)
 74         ret+=query(rch[rt],ll,rr,mid+1,r,add[rt]+add1);
 75     return ret;
 76 }
 77 int main(){
 78     int flag(0);
 79     while(scanf("%d%d",&n,&m)==2){
 80         if(flag)
 81             puts("");
 82         for(int i=1;i<=n;++i)
 83             v[i]=read();
 84         ++flag;
 85         cnt=t=0;
 86         rt[0]=build(1,n);
 87         while(m--){
 88             scanf("%s",op);
 89             if(op[0]==‘Q‘){
 90                 int x(read()),y(read());
 91                 printf("%lld\n",query(rt[t],x,y,1,n,0));
 92             }
 93             if(op[0]==‘C‘){
 94                 int x(read()),y(read());
 95                 L z(read());
 96                 rt[t+1]=update(rt[t],x,y,z,1,n);
 97                 ++t;
 98             }
 99             if(op[0]==‘H‘){
100                 int x(read()),y(read()),z(read());
101                 printf("%lld\n",query(rt[z],x,y,1,n,0));
102             }
103             if(op[0]==‘B‘){
104                 int x(read());
105                 t=x;
106                 cnt=rt[t+1]-1;
107             }
108         }
109     }
110 }

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