【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 ≤ 10 ⁵, |A _[i]| ≤ 10 ⁹, 1 ≤ l ≤ r ≤ N, |d| ≤ 10 ⁴ .. 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
A ₁ A ₂ ... 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 }

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