【HDU4348】【主席树】To the moon

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

【分析】

简单题，唯一要注意的是内存大小...

写指针各种被卡....无奈最后写数组...

看到有人用可持久化树状数组做...我开始算了一下好像会超内存，然后就没写了。。Orzzz

 /*
 唐代高蟾
 《金陵晚望》

 曾伴浮云归晚翠，犹陪落日泛秋声。
 世间无限丹青手，一片伤心画不成。
 */
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <utility>
 #include <iomanip>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <assert.h>
 #include <map>
 #include <ctime>
 #include <cstdlib>
 #include <stack>
 #define LOCAL
 const int MAXN =  + ;
 const int MAXM =  + ;
 const int INF = ;
 const int SIZE = ;
 const int maxnode =  0x7fffffff + ;
 using namespace std;
 typedef long long ll;
 int lson[MAXN], rson[MAXN];
 int data[MAXN], add[MAXN];
 ll sum[MAXN];
 int tot;

 int insert(int t, int ll, int rr, int d, int l, int r){
     int now = ++tot;
     lson[now] = lson[t];
     rson[now] = rson[t];
     add[now] = add[t];
     sum[now] = sum[t];
     sum[now] += (long long)(d * (rr - ll + ));
     if(ll == l && rr == r){
         add[now] += d;
         return now;
     }
     int mid = (l + r)>>;
     if(rr <= mid) lson[now] = insert(lson[t], ll, rr, d, l, mid);
     else if(ll > mid) rson[now] = insert(rson[t], ll, rr, d, mid + , r);
     else{
         lson[now] = insert(lson[t], ll, mid, d, l, mid);
         rson[now] = insert(rson[t], mid + , rr, d, mid + , r);
     }
     return now;
 }
 //在t的根内查询[ll, rr]区间的值
 long long query(int t, int ll, int rr, int l, int r){
    long long Ans = (long long)(add[t] * (rr - ll + ));
    if (ll == l && rr == r) return sum[t];
    int mid = (l + r)>>;
    if (rr <= mid) Ans += query(lson[t], ll, rr, l, mid);
    else if (ll > mid) Ans += query(rson[t], ll, rr, mid + , r);
    else {
         Ans += query(lson[t], ll, mid, l, mid);
         Ans += query(rson[t], mid + , rr, mid + , r);
    }
    return Ans;
 }
 int build(int ll, int rr){
     int now = ++tot;
     add[now] = ;
     if (ll == rr){
        scanf("%lld", &sum[now]);
        lson[now] = rson[now] = ;
        return now;
     }
     int mid = (ll + rr)>>;
     lson[now] = build(ll, mid);
     rson[now] = build(mid + , rr);
     sum[now] = sum[lson[now]] + sum[rson[now]];
     return now;
 }

 int n ,m;
 void work(){
      tot = ;
      data[] = build(,n);
      int now = ;
      for (int i = ; i <= m; i++){
          char str[];
          scanf("%s", str);
          if (str[] == 'Q'){
             int l, r;
             scanf("%d%d", &l, &r);
             printf("%lld\n", query(data[now], l, r, , n));
          }else if(str[] == 'C'){
                int l, r, d;
                scanf("%d%d%d", &l, &r, &d);
                data[now+] = insert(data[now], l, r, d, , n);
                now++;
          }else if(str[] == 'H'){
                int l, r, t;
                scanf("%d%d%d", &l, &r, &t);
                printf("%lld\n", query(data[t], l, r, , n));
          }else scanf("%d", &now);
      }
      printf("\n");
 }

 int main(){

     while (scanf("%d%d", &n, &m) != EOF){
           //scanf("%d%d", &n, &m);
           work();
     }
     return ;
 }