POJ - 3468 - A Simple Problem with Integers - 线段树Lazy大法

1.题目描述：

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 107104		Accepted: 33434
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

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

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

2.题意概述：

对数组A有两种操作Q x y是询问x,y区间的和C x y z是对[x, y]区间内所有数都加z

3.解题思路：

很容易想到用线段树维护，这里是区间更新，可以用lazy大法，简要的说，就是对于一个区间，如果它位于要更新的子区间内，那么就延缓对它儿子区间的更新，这样可以进一步降低update的复杂度。

4.AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define maxn 100010
#define lson root << 1
#define rson root << 1 | 1
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
struct tree
{
	int l, r;
	ll w, lazy;
} t[maxn << 2];
void pushup(int root)
{
	t[root].w = t[lson].w + t[rson].w;
}
void pushdown(int root)
{
	if (t[root].lazy)
	{
		t[lson].lazy += t[root].lazy;
		t[lson].w += (t[lson].r - t[lson].l + 1) * t[root].lazy;
		t[rson].lazy += t[root].lazy;
		t[rson].w += (t[rson].r - t[rson].l + 1) * t[root].lazy;
		t[root].lazy = 0;
	}
}
void build(int l, int r, int root)
{
	t[root].l = l;
	t[root].r = r;
	t[root].lazy = 0;
	if (l == r)
	{
		scanf("%lld", &t[root].w);
		return;
	}
	int mid = l + r >> 1;
	build(l, mid, lson);
	build(mid + 1, r, rson);
	pushup(root);
}
void update(ll c, int l, int r, int root)
{
	if (l <= t[root].l && t[root].r <= r)
	{
		t[root].lazy += c;
		t[root].w += c * (t[root].r - t[root].l + 1);
		return;
	}
	pushdown(root);
	int mid = t[root].l + t[root].r >> 1;
	if (l <= mid)
		update(c, l, r, lson);
	if (r > mid)
		update(c, l, r, rson);
	pushup(root);
}
ll query(int l, int r, int root)
{
	if (l <= t[root].l && t[root].r <= r)
		return t[root].w;
	pushdown(root);
	int mid = t[root].l + t[root].r >> 1;
	ll ans = 0;
	if (l <= mid)
		ans += query(l, r, lson);
	if (r > mid)
		ans += query(l, r, rson);
	return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	long _begin_time = clock();
#endif
	int n, q;
	char op[3];
	while (~scanf("%d%d", &n, &q))
	{
		build(1, n, 1);
		while (q--)
		{
			int a, b;
			ll c;
			scanf("%s%d%d", op, &a, &b);
			if (op[0] == 'Q')
				printf("%lld\n", query(a, b, 1));
			else
			{
				scanf("%lld", &c);
				update(c, a, b, 1);
			}
		}
	}
#ifndef ONLINE_JUDGE
	long _end_time = clock();
	printf("time = %ld ms.", _end_time - _begin_time);
#endif
	return 0;
}