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

标签：带修主席树

题目大意：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。

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。

直接上代码：

#include <iostream>#include <cstdio>#define MAX_N 100000using 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;}