hihocoder 1080 更为复杂的买卖房屋姿势线段树

最新推荐文章于 2018-10-28 17:11:00 发布

飞火扑蛾

最新推荐文章于 2018-10-28 17:11:00 发布

阅读量836

点赞数

CC 4.0 BY-SA版权

分类专栏：线段树文章标签： hihocoder 线段树

本文链接：https://blog.youkuaiyun.com/u010535824/article/details/41852413

线段树专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种解决数组更新问题的方法，通过构建线段树实现两种更新操作：加法更新与赋值更新，并提供了完整的C++代码示例。适用于需要高效处理区间更新与查询的应用场景。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

传送门:hihocoder 1080

给定一个长度为n的数组，有两种操作：

(1)对范围[l,r]内的数都加上x(x可为负)

(2)把范围[l,r]内的数都变成x

问每一步操作后整个数组的和为多少

两种更新姿势的线段树,写好对应的更新函数即可

/******************************************************
 * File Name:   building.cpp
 * Author:      kojimai
 * Create Time: 2014年12月01日 星期一 20时48分38秒
 ******************************************************/

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
const int FFF = 100005;
struct nodes
{
	int l,r,sum,add,change;
}node[FFF<<2];
void build(int l,int r,int num)
{
	node[num].l = l; node[num].r = r; node[num].add = 0; node[num].change = 0;
	if(l == r)
	{
		scanf("%d",&node[num].sum);
		return;
	}
	int mid = (l+r) >> 1;
	build(l,mid,num<<1);
	build(mid+1,r,num<<1|1);
	node[num].sum = node[num<<1].sum + node[num<<1|1].sum;
	return;
}

void push_change(int num)
{
	node[num<<1].change = node[num].change;
	node[num<<1].add = 0;
	node[num<<1].sum = (node[num<<1].r - node[num<<1].l + 1) * node[num<<1].change;
	node[num<<1|1].change = node[num].change;
	node[num<<1|1].add = 0;
	node[num<<1|1].sum = (node[num<<1|1].r - node[num<<1|1].l + 1) * node[num<<1|1].change;
	node[num].change = 0;
}
void push_add(int num)
{
	node[num<<1].add += node[num].add;
	node[num<<1].sum += (node[num<<1].r - node[num<<1].l + 1) * node[num].add;
	node[num<<1|1].add += node[num].add;
	node[num<<1|1].sum += (node[num<<1|1].r - node[num<<1|1].l + 1)* node[num].add;
	node[num].add = 0;
}

void update1(int l,int r,int z,int num)
{
	if(node[num].l == node[num].r)
	{
		node[num].sum = z;
		node[num].add = node[num].change = 0;
		return;
	}
	if(node[num].change)
		push_change(num);
	if(node[num].add)
		push_add(num);
	if(node[num].l == l && node[num].r == r)
	{
		node[num].change = z;
		node[num].sum = (r-l+1)*z;
		node[num].add = 0;
		return;
	}
	int mid = (node[num].l + node[num].r) >> 1;
	if(mid >= r)
		update1(l,r,z,num<<1);
	else if(mid < l)
		update1(l,r,z,num<<1|1);
	else
	{
		update1(l,mid,z,num<<1);
		update1(mid+1,r,z,num<<1|1);
	}
	node[num].sum = node[num<<1].sum + node[num<<1|1].sum;
	return;
}

void update0(int l,int r,int z,int num)
{
	if(node[num].l == node[num].r)
	{
		node[num].sum += z;
		node[num].add = node[num].change = 0;
		//cout<<"num = "<<num<<" nl = "<<node[num].l<<" nr = "<<node[num].r<<" l = "<<l<<" r = "<<r<<" sum = "<<node[num].sum<<endl;
		return;
	}
	if(node[num].change)
		push_change(num);
	if(node[num].add)
		push_add(num);
	if(node[num].l == l && node[num].r == r)
	{
		node[num].change = 0;
		node[num].sum += z * (r - l + 1);
		node[num].add = z;
		//cout<<"num = "<<num<<" nl = "<<node[num].l<<" nr = "<<node[num].r<<" l = "<<l<<" r = "<<r<<" sum = "<<node[num].sum<<endl;
		return;
	}
	int mid = (node[num].l + node[num].r) >> 1;
	if(mid >= r)
		update0(l,r,z,num<<1);
	else if(mid < l)
		update0(l,r,z,num<<1|1);
	else
	{
		update0(l,mid,z,num<<1);
		update0(mid+1,r,z,num<<1|1);
	}
	node[num].sum = node[num<<1].sum + node[num<<1|1].sum;
	//cout<<"num = "<<num<<" nl = "<<node[num].l<<" nr = "<<node[num].r<<" l = "<<l<<" r = "<<r<<" sum = "<<node[num].sum<<endl;
	return;
}

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	build(1,n+1,1);

	int type,x,y,z;
	while(m--)
	{
		scanf("%d%d%d%d",&type,&x,&y,&z);
		if(type == 1)
			update1(x+1,y+1,z,1);
		else
			update0(x+1,y+1,z,1);
		printf("%d\n",node[1].sum);
	}
	return 0;
}