HDU 4348 SPOJ 11470 To the moon

Vjudge题面

Time limit 2000 ms
Memory limit 65536 kB
OS Windows
Source 2012 Multi-University Training Contest 5

SPOJ原版题面

Background

To The Moon is a independent game released in November 2011, it is a role-playing adventure game powered by RPG Maker.

River && Anya .. .

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.

Description

You‘ve been given N integers A[1], A[2],..., A[N]. On these integers, you need to implement the following operations:

  • C l r d: Adding a constant d for every {Ai | l <= i <= r}, and increase the timestamp by 1, this is the only operation that will cause the timestamp increase.
  • Q l r: Querying the current sum of {Ai | l <= i <= r}.
  • H l r t: Querying a history sum of {Ai | l <= i <= r} in time t.
  • 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
A1 A2 ... An... 
(here following the m operations. )

Output

... (for each query, simply print the result. )

Example

Input 1:

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

Output 1:

4
55
9
15

Input 2:

2 4
0 0
C 1 1 1
C 2 2 -1
Q 1 2
H 1 2 1

Output 2:

0
1

吐槽

样例有点弱……传参时\(mid+1\)\(r\)位置写反了,还过了样例,debug好久……
另外,HDU(连带着vjudge)上的样例是假的,各个case的输出之间不用换行的。换行会PE。

解题思路

题意就是写记录历史版本的线段树,也就是可持久化线段树,那就直接把区间修改的线段树板子拿过来改吧。但写着写着发现lazy标记下放会导致MLE,于是查阅资料,找到解决方案——标记永久化。lazy标记在区间加时不下放,而在区间查询时一路向下累加lazy值,当该区间被全部包含在查询区间中时,把当前区间的sum加上lazy的贡献相加,返回上去。

洛谷题解区看见另一种据说更优的方案——
[SP11470]TTM-To the moon——主席树区间修改/差分主席树里面的法二,用了两棵主席树,没太看懂,先留坑不填了

源代码

#include <stdio.h>

int T;
int n, m;
long long a[100010];

struct Node
{
	int lson, rson;
	long long sum, lazy;
} t[4000010];
int cnt = 1, root[100010], timer;
void build(int &x, int l, int r)
{
	x = cnt++;
	t[x].lazy = 0;
	if (l == r)
	{
		t[x].sum = a[l];
		return;
	}
	int mid = l + r >> 1;
	build(t[x].lson, l, mid);
	build(t[x].rson, mid + 1, r);
	t[x].sum = t[t[x].lson].sum + t[t[x].rson].sum;
}

void add(int &x, int last, int l, int r, int al, int ar, int d)
{
	if (al > r || ar < l)
		return;
	x = cnt++;
	t[x] = t[last];
	if (al <= l && r <= ar)
	{
		t[x].lazy += d;
		t[x].sum += (r - l + 1) * d;
		return;
	}
	int mid = l + r >> 1;
	add(t[x].lson, t[last].lson, l, mid, al, ar, d);
	add(t[x].rson, t[last].rson, mid + 1,r, al, ar, d);
	t[x].sum = t[t[x].lson].sum + t[t[x].rson].sum + t[x].lazy * (r - l + 1);
}

long long que(int x, int l, int r, int ql, int qr, long long la)
{
	if (ql > r || qr < l)
		return 0;
	if (ql <= l && r <= qr)
		return t[x].sum + la * (r - l + 1);
	la += t[x].lazy;
	int mid = l + r >> 1;
	return que(t[x].lson, l, mid, ql, qr, la) + que(t[x].rson, mid + 1, r, ql, qr, la);
}

int main()
{
	freopen("test.in", "r", stdin);
	//scanf("%d", &T);
	//T=1;
	//while (T--)
	while (~scanf("%d%d", &n, &m))
	{
		cnt = 1;
		timer = 0;
		for (int i = 1; i <= n; i++)
			scanf("%lld", a + i);
		build(root[0], 1, n);
		char opt[2];
		int l, r, t;
		long long d;
		while (m--)
		{
			scanf("%s", opt);
			if (opt[0] == 'C')
			{
				scanf("%d%d%lld", &l, &r, &d);
				add(root[timer + 1], root[timer], 1, n, l, r, d);
				timer++;
			}
			else if (opt[0] == 'Q')
			{
				scanf("%d%d", &l, &r);
				printf("%lld\n", que(root[timer], 1, n, l, r, 0));
			}
			else if (opt[0] == 'H')
			{
				scanf("%d%d%d", &l, &r, &t);
				printf("%lld\n", que(root[t], 1, n, l, r, 0));
			}
			else
			{
				scanf("%d", &t);
				timer = t;
			}
		}
		//puts("");//hdu样例假了,这句不能要
	}
	return 0;
}
posted @ 2019-07-25 15:23  wawcac  阅读(193)  评论(0编辑  收藏  举报