# HDU4348 To The Moon <带修主席树>

【HDU4348】To The Moon

Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)

Problem 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 {Ai | 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 {Ai | l <= i <= r}.
3. H l r t: Querying a history sum of {Ai | 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 ... 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

#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;
}

posted @ 2017-09-20 15:34  Azrael_Death  阅读(200)  评论(0编辑  收藏  举报