【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 = 3000000 + 10;
const int MAXM = 1000000 + 10;
const int INF = 100000000;
const int SIZE = 450;
const int maxnode =  0x7fffffff + 10;
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 + 1));
    if(ll == l && rr == r){
        add[now] += d;
        return now;
    }
    int mid = (l + r)>>1;
    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 + 1, r);
    else{
        lson[now] = insert(lson[t], ll, mid, d, l, mid);
        rson[now] = insert(rson[t], mid + 1, rr, d, mid + 1, 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 + 1));
   if (ll == l && rr == r) return sum[t];
   int mid = (l + r)>>1;
   if (rr <= mid) Ans += query(lson[t], ll, rr, l, mid);
   else if (ll > mid) Ans += query(rson[t], ll, rr, mid + 1, r);
   else {
        Ans += query(lson[t], ll, mid, l, mid);
        Ans += query(rson[t], mid + 1, rr, mid + 1, r);
   }
   return Ans;
}
int build(int ll, int rr){
    int now = ++tot;
    add[now] = 0;
    if (ll == rr){
       scanf("%lld", &sum[now]);
       lson[now] = rson[now] = 0;
       return now;
    }
    int mid = (ll + rr)>>1;
    lson[now] = build(ll, mid);
    rson[now] = build(mid + 1, rr);
    sum[now] = sum[lson[now]] + sum[rson[now]];
    return now;
}

int n ,m;
void work(){
     tot = 0;
     data[0] = build(1,n);
     int now = 0;
     for (int i = 1; i <= m; i++){
         char str[3];
         scanf("%s", str);
         if (str[0] == 'Q'){
            int l, r;
            scanf("%d%d", &l, &r);
            printf("%lld
", query(data[now], l, r, 1, n));
         }else if(str[0] == 'C'){
               int l, r, d;
               scanf("%d%d%d", &l, &r, &d);
               data[now+1] = insert(data[now], l, r, d, 1, n);
               now++;
         }else if(str[0] == 'H'){
               int l, r, t;
               scanf("%d%d%d", &l, &r, &t);
               printf("%lld
", query(data[t], l, r, 1, n));
         }else scanf("%d", &now);
     }
     printf("
");
}

int main(){

    while (scanf("%d%d", &n, &m) != EOF){
          //scanf("%d%d", &n, &m);
          work(); 
    }
    return 0;
}