【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

  1 /*
  2 唐代高蟾
  3 《金陵晚望》
  4
  5 曾伴浮云归晚翠，犹陪落日泛秋声。
  6 世间无限丹青手，一片伤心画不成。
  7 */
  8 #include <iostream>
  9 #include <cstdio>
 10 #include <algorithm>
 11 #include <cstring>
 12 #include <vector>
 13 #include <utility>
 14 #include <iomanip>
 15 #include <string>
 16 #include <cmath>
 17 #include <queue>
 18 #include <assert.h>
 19 #include <map>
 20 #include <ctime>
 21 #include <cstdlib>
 22 #include <stack>
 23 #define LOCAL
 24 const int MAXN = 3000000 + 10;
 25 const int MAXM = 1000000 + 10;
 26 const int INF = 100000000;
 27 const int SIZE = 450;
 28 const int maxnode =  0x7fffffff + 10;
 29 using namespace std;
 30 typedef long long ll;
 31 int lson[MAXN], rson[MAXN];
 32 int data[MAXN], add[MAXN];
 33 ll sum[MAXN];
 34 int tot;
 35
 36
 37 int insert(int t, int ll, int rr, int d, int l, int r){
 38     int now = ++tot;
 39     lson[now] = lson[t];
 40     rson[now] = rson[t];
 41     add[now] = add[t];
 42     sum[now] = sum[t];
 43     sum[now] += (long long)(d * (rr - ll + 1));
 44     if(ll == l && rr == r){
 45         add[now] += d;
 46         return now;
 47     }
 48     int mid = (l + r)>>1;
 49     if(rr <= mid) lson[now] = insert(lson[t], ll, rr, d, l, mid);
 50     else if(ll > mid) rson[now] = insert(rson[t], ll, rr, d, mid + 1, r);
 51     else{
 52         lson[now] = insert(lson[t], ll, mid, d, l, mid);
 53         rson[now] = insert(rson[t], mid + 1, rr, d, mid + 1, r);
 54     }
 55     return now;
 56 }
 57 //在t的根内查询[ll, rr]区间的值
 58 long long query(int t, int ll, int rr, int l, int r){
 59    long long Ans = (long long)(add[t] * (rr - ll + 1));
 60    if (ll == l && rr == r) return sum[t];
 61    int mid = (l + r)>>1;
 62    if (rr <= mid) Ans += query(lson[t], ll, rr, l, mid);
 63    else if (ll > mid) Ans += query(rson[t], ll, rr, mid + 1, r);
 64    else {
 65         Ans += query(lson[t], ll, mid, l, mid);
 66         Ans += query(rson[t], mid + 1, rr, mid + 1, r);
 67    }
 68    return Ans;
 69 }
 70 int build(int ll, int rr){
 71     int now = ++tot;
 72     add[now] = 0;
 73     if (ll == rr){
 74        scanf("%lld", &sum[now]);
 75        lson[now] = rson[now] = 0;
 76        return now;
 77     }
 78     int mid = (ll + rr)>>1;
 79     lson[now] = build(ll, mid);
 80     rson[now] = build(mid + 1, rr);
 81     sum[now] = sum[lson[now]] + sum[rson[now]];
 82     return now;
 83 }
 84
 85 int n ,m;
 86 void work(){
 87      tot = 0;
 88      data[0] = build(1,n);
 89      int now = 0;
 90      for (int i = 1; i <= m; i++){
 91          char str[3];
 92          scanf("%s", str);
 93          if (str[0] == ‘Q‘){
 94             int l, r;
 95             scanf("%d%d", &l, &r);
 96             printf("%lld\n", query(data[now], l, r, 1, n));
 97          }else if(str[0] == ‘C‘){
 98                int l, r, d;
 99                scanf("%d%d%d", &l, &r, &d);
100                data[now+1] = insert(data[now], l, r, d, 1, n);
101                now++;
102          }else if(str[0] == ‘H‘){
103                int l, r, t;
104                scanf("%d%d%d", &l, &r, &t);
105                printf("%lld\n", query(data[t], l, r, 1, n));
106          }else scanf("%d", &now);
107      }
108      printf("\n");
109 }
110
111 int main(){
112
113     while (scanf("%d%d", &n, &m) != EOF){
114           //scanf("%d%d", &n, &m);
115           work();
116     }
117     return 0;
118 }