poj3468:A Simple Problem with Integers（线段树lazy）

最新推荐文章于 2022-12-09 12:45:57 发布

Acc_plus

最新推荐文章于 2022-12-09 12:45:57 发布

阅读量242

点赞数

CC 4.0 BY-SA版权

分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/qq_35371796/article/details/78384119

线段树专栏收录该内容

1 篇文章

订阅专栏

本文介绍了解决POJ3468问题的方法，采用线段树区间更新算法，并利用lazy标签优化效率。文章详细说明了节点结构，包括区间值、lazy标签及区间长度等关键元素。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

poj3468 :A Simple Problem with Integers

普通的线段树区间更新，打上lazy标签即可，一个节点记录三个值，一个是区间的值，一个是lazy标签，记录了还未往下传递的值，另外一个是这个区间的长度（长度乘以传下来的lazy即是这个区间增加的值，建树的时候可以算出）。注意开好long long即可。

#include
  
   
#include
   
    
#include
    
     
#include
     
      
#include
      
       
#define MAXN 400000
using namespace std;
struct node
{
	long long value;
	long long lazy; 
	int len;
};
node tree[MAXN << 1]{};
long long value[MAXN];
int left_, right_;
int length;
long long buildtree(int index, int left, int right)
{
	tree[index].len = right - left + 1;
	if (left == right)
	{
		tree[index].value = value[left];
	}
	else
	{
		int mid = (left + right) >> 1;
		tree[index].value = buildtree(index << 1, left, mid)
			+ buildtree(index << 1 | 1, mid + 1, right);
	}
	return tree[index].value;
}
long long getvalue(int index, int templeft, int tempright)
{
	int lson = index << 1;
	int rson = index << 1 | 1;
	int mid = (templeft + tempright) >> 1;
	if (templeft >= left_ && tempright <= right_)
		return tree[index].value;
	if (templeft > right_ || tempright < left_)
		return 0;
	if (tree[index].lazy)
	{
		long long &lazy = tree[index].lazy;
		tree[lson].lazy += lazy;
		tree[lson].value += lazy * tree[lson].len;
		tree[rson].lazy += lazy;
		tree[rson].value += lazy * tree[rson].len;
		lazy = 0;
	}
	return getvalue(lson, templeft, mid) + getvalue(rson, mid + 1, tempright);
}
void modify(int index, long long value, int templeft, int tempright)
{
	if (templeft >= left_ && tempright <= right_)
	{
		tree[index].lazy += value;
		tree[index].value += value*tree[index].len;
		return;
	}
	if (templeft > right_ || tempright < left_)
		return;
	int mid = (templeft + tempright) >> 1;
	int lson = index << 1;
	int rson = index << 1 | 1;
	if (tree[index].lazy)
	{
		long long &lazy = tree[index].lazy;
		tree[lson].lazy += lazy;
		tree[lson].value += lazy * tree[lson].len;
		tree[rson].lazy += lazy;
		tree[rson].value += lazy * tree[rson].len;
		lazy = 0;
	}
	modify(lson, value, templeft, mid);
	modify(rson, value, mid + 1, tempright);
	tree[index].value = tree[lson].value + tree[rson].value;
}

int main()
{
	int size, opr;
	scanf("%d%d", &size, &opr);
	for (int i = 1; i <= size; i++)
		scanf("%lld", &value[i]);
	buildtree(1, 1, size);
	char ch;
	int a, b;
	long long c;
	getchar();
	while (opr--)
	{
		ch = getchar();
		if (ch == 'Q')
		{
			scanf("%d%d", &a, &b);
			left_ = a; right_ = b;
			printf("%lld\n", getvalue(1, 1, size));
		}
		else
		{
			scanf("%d%d%lld", &a, &b, &c);
			left_ = a; right_ = b;
			modify(1, c, 1, size);
		}
		getchar();
	}
}