HDU4348 To The Moon <带修主席树>

Problem

【HDU4348】To The Moon

$\mathrm{Time\;Limit:\;4000/2000\;MS\;(Java/Others)}$
$\mathrm{Memory\;Limit:\;65536/65536\;KB\;(Java/Others)}$

Description

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 $\{A_i | l \le i \le r\}$, 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 $\{A_i | l \le i \le r\}$.
3. $H$ $l$ $r$ $t$: Querying a history sum of $\{A_i | l \le i \le r\}$ 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.

$N, M \le 10^5$, $|A[i]|\le 10^9$, $1\le l\le r\le N$, $|d|\le 10^4$. The system start from time $0$, and the first modification is in time $1$, $t\ge 0$, 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

标签：带修主席树

Translation

维护一个数据结构，使得其可有四种操作：区间修改，区间求和，某时间的区间和，返回某时间。

Solution

很明显这是一道主席树的板子题。不过此题要带区间修改。
普通区间修改需要加$tag$，并不断下传。而对于主席树，下传意味着新建节点，可能会$MLE$。所以这里我们暴力一点，直接标记永久化，这样写起来简洁，而且省空间。

Code

#include <iostream>
#include <cstdio>
#define MAX_N 100000
using namespace std;
struct node {int ls, rs; long long val, tag;} tr[MAX_N*50+5];
int n, m;
int cnt, now, root[MAX_N+5];
void updata(int v, int s, int t) {tr[v].val = tr[tr[v].ls].val+tr[tr[v].rs].val+(long long)(t-s+1)*tr[v].tag;}
void build(int v, int s, int t) {
    tr[v].ls = tr[v].rs = tr[v].tag = tr[v].val = 0;
    if (s == t) {
        scanf("%I64d", &tr[v].val);
        return;
    }
    tr[v].ls = ++cnt, tr[v].rs = ++cnt;
    int mid = s+t>>1;
    build(tr[v].ls, s, mid);
    build(tr[v].rs, mid+1, t);
    updata(v, s, t);
}
void modify(int v, int o, int s, int t, int l, int r, long long x) {
    tr[v] = tr[o];
    if (s >= l && t <= r) {
        tr[v].tag += x;
        tr[v].val += (long long)(t-s+1)*x;
        return;
    }
    int mid = s+t>>1;
    if (l <= mid)	modify(tr[v].ls = ++cnt, tr[o].ls, s, mid, l, r, x);
    if (r >= mid+1)	modify(tr[v].rs = ++cnt, tr[o].rs, mid+1, t, l, r, x);
    updata(v, s, t);
}
long long query(int v, int s, int t, int l, int r, long long tot) {
    if (s >= l && t <= r)	return tr[v].val+(long long)(t-s+1)*tot;
    tot += tr[v].tag;
    int mid = s+t>>1;
    long long ret = 0;
    if (l <= mid)	ret += query(tr[v].ls, s, mid, l, r, tot);
    if (r >= mid+1)	ret += query(tr[v].rs, mid+1, t, l, r, tot);
    return ret;
}
int main() {
    while(scanf("%d%d", &n, &m) != EOF) {
        cnt = now = 0;
        root[now] = ++cnt;
        build(root[now], 1, n);
        while (m--) {
            char ch;
            cin >> ch;
            if (ch == 'C') {
                int l, r;
                long long d;
                scanf("%d%d%I64d", &l, &r, &d);
                now++;
                root[now] = ++cnt;
                modify(root[now], root[now-1], 1, n, l, r, d);
            }
            if (ch == 'Q') {
                int l, r;
                scanf("%d%d", &l, &r);
                printf("%I64d\n", query(root[now], 1, n, l, r, 0LL));
            }
            if (ch == 'H') {
                int l, r, t;
                scanf("%d%d%d", &l, &r, &t);
                printf("%I64d\n", query(root[t], 1, n, l, r, 0LL));
            }
            if (ch == 'B') {
                int t;
                scanf("%d", &t);
                now = t;
            }
        }
    }
    return 0;
}