Codeforces 295A Greg and Array 线段树区间累加，单点询问

本文链接：https://blog.youkuaiyun.com/w703710691d/article/details/42970423

本文介绍了一道算法题目，通过使用线段树解决区间更新问题。具体地，文章详细阐述了如何通过两次构建线段树，分别计算每个操作的执行次数及最终数组的值。

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A. Greg and Array

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Greg has an array a = a₁, a₂, ..., a_n and m operations. Each operation looks as: l_i, r_i, d_i, (1 ≤ l_i ≤ r_i ≤ n). To apply operation i to the array means to increase all array elements with numbers l_i, l_i + 1, ..., r_i by value d_i.

Greg wrote down k queries on a piece of paper. Each query has the following form: x_i, y_i, (1 ≤ x_i ≤ y_i ≤ m). That means that one should apply operations with numbers x_i, x_i + 1, ..., y_i to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input

The first line contains integers n, m, k (1 ≤ n, m, k ≤ 10⁵). The second line contains n integers: a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁵) — the initial array.

Next m lines contain operations, the operation number i is written as three integers: l_i, r_i, d_i, (1 ≤ l_i ≤ r_i ≤ n), (0 ≤ d_i ≤ 10⁵).

Next k lines contain the queries, the query number i is written as two integers: x_i, y_i, (1 ≤ x_i ≤ y_i ≤ m).

The numbers in the lines are separated by single spaces.

Output

On a single line print n integers a₁, a₂, ..., a_n — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Sample test(s)

Input

Output

9 18 17

Input

Output

Input

Output

5 18 31 20

题意：给出一个数列，长度为n，m个操作，k个对操作的执行。对于每一个操作，将数列的区间[Li，Ri]都加上Di。对于每一个执行，对操作[Xi,Yi]都执行一次。最后输出所有执行之后，这个数列的每一个元素的值。

思路：计算出每一个操作执行的次数，也就是每一个操作总共会累加多少值，在对数列进行区间累加即可。要算出每个操作执行了多少次，可以用线段树对操作的下标进行建树，对于每一个执行，对这个线段树进行区间累加1的操作。这样就能算出每个操作执行了多少次，也就可以算出，每个操作最后累加了多少。最后再用另一个线段树维护数列即可

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#define LL long long
using namespace std;
struct node
{
	int l, r;
	LL add;
}tree[400005];
void pushdown(int id)
{
	if (tree[id].add)
	{
		tree[id << 1].add += tree[id].add;
		tree[id << 1 | 1].add += tree[id].add;
		tree[id].add = 0;
	}
}
void build(int id, int l, int r)
{
	tree[id].l = l;
	tree[id].r = r;
	tree[id].add = 0;
	if (l != r)
	{
		int mid = (l + r) >> 1;
		build(id << 1, l, mid);
		build(id << 1 | 1, mid + 1, r);
	}
}
void op(int id, int l, int r, LL add)
{
	if (l <= tree[id].l&&tree[id].r <= r)
	{
		tree[id].add += add;
	}
	else
	{
		pushdown(id);
		int mid = (tree[id].l + tree[id].r) >> 1;
		if (l <= mid) op(id << 1, l, r, add);
		if (mid<r) op(id << 1 | 1, l, r, add);
	}
}
int a[100005];
int b[100005];
struct operate
{
	int l, r;
	LL add;
}OP[100005];
void out(int id)
{
	if (tree[id].l == tree[id].r)
	{
		b[tree[id].l] = tree[id].add;
	}
	else
	{
		pushdown(id);
		out(id << 1);
		out(id << 1 | 1);
	}
}
void out2(int id)
{
	if (tree[id].l == tree[id].r)
	{
		printf("%I64d ", a[tree[id].l] + tree[id].add);
	}
	else
	{
		pushdown(id);
		out2(id << 1);
		out2(id << 1 | 1);
	}
}
int main()
{
	int n, m, k;
	scanf("%d %d %d", &n, &m, &k);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	build(1, 1, m);
	for (int i = 1; i <= m; i++)
	{
		scanf("%d %d %I64d", &OP[i].l, &OP[i].r, &OP[i].add);
	}
	while (k--)
	{
		int x, y;
		scanf("%d %d", &x, &y);
		op(1, x, y, 1);
	}
	out(1);
	build(1, 1, n);
	for (int i = 1; i <= m; i++)
	{
		if (!b[i]) continue;
		op(1, OP[i].l, OP[i].r, OP[i].add*b[i]);
	}
	out2(1);
	return 0;
	
}