• 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 {Ai | 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 {Ai | l <= i <= r}.
    3. H l r t: Querying a history sum of {Ai | 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 ... An
    ... (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
     
     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。
     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 
     5 #define MP make_pair
     6 #define PB push_back
     7 typedef long long LL;
     8 typedef pair<int,int> PII;
     9 const double eps=1e-8;
    10 const double pi=acos(-1.0);
    11 const int K=5e6+7;
    12 const int mod=1e9+7;
    13 
    14 int tot,ls[K],rs[K],rt[K];
    15 LL sum[K],add[K],v[K];
    16 //sum[o]记录的是该节点区间内出现的数的个数
    17 //区间指的是将数离散化后的区间
    18 void build(int &o,int l,int r)
    19 {
    20     o=tot++,sum[o]=v[l],add[o]=0;
    21     if(l==r) return ;
    22     int mid=l+r>>1;
    23     build(ls[o],l,mid),build(rs[o],mid+1,r);
    24     sum[o]=sum[ls[o]]+sum[rs[o]];
    25 }
    26 void update(int &o,int p,int l,int r,int nl,int nr,LL x)
    27 {
    28     o=tot++,ls[o]=ls[p],rs[o]=rs[p],sum[o]=sum[p]+(nr-nl+1)*x,add[o]=add[p];
    29     if(l==nl && r==nr){add[o]+=x;return;}
    30     int mid=l+r>>1;
    31     if(nr<=mid) update(ls[o],ls[p],l,mid,nl,nr,x);
    32     else if(nl>mid) update(rs[o],rs[p],mid+1,r,nl,nr,x);
    33     else update(ls[o],ls[p],l,mid,nl,mid,x),update(rs[o],rs[p],mid+1,r,mid+1,nr,x);
    34     sum[o]=sum[ls[o]]+sum[rs[o]]+(r-l+1)*add[o];
    35 }
    36 LL query(int o,int l,int r,int nl,int nr,LL f)
    37 {
    38     if(l==nl && r==nr) return sum[o]+(nr-nl+1)*f;
    39     int mid=l+r>>1;
    40     if(nr<=mid) return query(ls[o],l,mid,nl,nr,f+add[o]);
    41     else if(nl>mid) return query(rs[o],mid+1,r,nl,nr,f+add[o]);
    42     return query(ls[o],l,mid,nl,mid,f+add[o])+query(rs[o],mid+1,r,mid+1,nr,f+add[o]);
    43 }
    44 int main(void)
    45 {
    46     int n,m;
    47     while(~scanf("%d%d",&n,&m))
    48     {
    49         tot=0;
    50         for(int i=1;i<=n;i++)
    51             scanf("%I64d",v+i);
    52         build(rt[0],1,n);
    53         int k=0,l,r,d;
    54         char op[10];
    55         while(m--)
    56         {
    57             scanf("%s",op);
    58             if(op[0]=='C')
    59                 scanf("%d%d%d",&l,&r,&d),update(rt[k+1],rt[k],1,n,l,r,d),k++;
    60             else if(op[0]=='Q')
    61                 scanf("%d%d",&l,&r),printf("%I64d
    ",query(rt[k],1,n,l,r,0));
    62             else if(op[0]=='H')
    63                 scanf("%d%d%d",&l,&r,&d),printf("%I64d
    ",query(rt[d],1,n,l,r,0));
    64             else
    65                 scanf("%d",&k);
    66         }
    67     }
    68     return 0;
    69 }
  • 相关阅读:
    docker学习
    LIS是什么?
    Android网络课程笔记-----定制通知系统
    java常用算法
    java正则表达式大全
    无需ROOT就能拿ANR日志
    Android网络课程笔记-----定制系统控件2
    Android网络课程笔记-----定制系统控件1
    Android网络课程笔记-----Fragment
    读《启示录》有感-----1
  • 原文地址:https://www.cnblogs.com/weeping/p/6924076.html
Copyright © 2020-2023  润新知