SPOJ - HORRIBLE 【线段树】
思路
线段树 区间更新 模板题 注意数据范围
AC代码
#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
using namespace std;
typedef long long LL;
const double PI = 3.14159265358979323846264338327;
const double E = 2.718281828459;
const double eps = 1e-6;
const int MAXN = 0x3f3f3f3f;
const int MINN = 0xc0c0c0c0;
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;
LL t, n, q;
LL anssum;
struct Node
{
LL l, r;
LL addv, sum;
}tree[maxn << 2];
void maintain(LL id)
{
if (tree[id].l >= tree[id].r)
return;
tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum;
}
void pushdown(LL id)
{
if (tree[id].l >= tree[id].r)
return;
if (tree[id].addv)
{
LL tmp = tree[id].addv;
tree[id << 1].addv += tmp;
tree[id << 1 | 1].addv += tmp;
tree[id << 1].sum += (tree[id << 1].r - tree[id << 1].l + 1) * tmp;
tree[id << 1 | 1].sum += (tree[id << 1 | 1].r - tree[id << 1 | 1].l + 1) * tmp;
tree[id].addv = 0;
}
}
void build(LL id, LL l, LL r)
{
tree[id].l = l;
tree[id].r = r;
tree[id].addv = 0;
tree[id].sum = 0;
if (l == r)
{
tree[id].sum = 0;
return;
}
LL mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
maintain(id);
}
void updateAdd(LL id, LL l, LL r, LL val)
{
if (tree[id].l >= l && tree[id].r <= r)
{
tree[id].addv += val;
tree[id].sum += (tree[id].r - tree[id].l + 1) * val;
return;
}
pushdown(id);
LL mid = (tree[id].l + tree[id].r) >> 1;
if (l <= mid)
updateAdd(id << 1, l, r, val);
if (mid < r)
updateAdd(id << 1 | 1, l, r, val);
maintain(id);
}
void query(LL id, LL l, LL r)
{
if (tree[id].l >= l && tree[id].r <= r)
{
anssum += tree[id].sum;
return;
}
pushdown(id);
LL mid = (tree[id].l + tree[id].r) >> 1;
if (l <= mid)
query(id << 1, l, r);
if (mid < r)
query(id << 1 | 1, l, r);
maintain(id);
}
int main()
{
scanf("%lld", &t);
while (t--)
{
scanf("%lld %lld", &n, &q);
build(1, 1, n);
LL id;
LL x, y;
LL val;
while (q--)
{
scanf("%lld", &id);
if (id == 0)
{
scanf("%lld %lld %lld", &x, &y, &val);
updateAdd(1, x, y, val);
}
else if (id == 1)
{
scanf("%lld %lld", &x, &y);
anssum = 0;
query(1, x, y);
cout << anssum << endl;
}
}
}
}
SPOJ - HORRIBLE 【线段树】的更多相关文章
- SPOJ GSS3 线段树系列1
SPOJ GSS系列真是有毒啊! 立志刷完,把线段树搞完! 来自lydrainbowcat线段树上的一道例题.(所以解法参考了lyd老师) 题意翻译 n 个数, q 次操作 操作0 x y把 Ax 修 ...
- SPOJ - GSS1 —— 线段树 (结点信息合并)
题目链接:https://vjudge.net/problem/SPOJ-GSS1 GSS1 - Can you answer these queries I #tree You are given ...
- SPOJ 2713 线段树(sqrt)
题意: 给你n个数(n <= 100000),然后两种操作,0 x y :把x-y的数全都sqrt ,1 x y:输出 x-y的和. 思路: 直接线段树更新就行了,对于当 ...
- SPOJ COT3 Combat on a tree(Trie树、线段树的合并)
题目链接:http://www.spoj.com/problems/COT3/ Alice and Bob are playing a game on a tree of n nodes.Each n ...
- SPOJ 2916 Can you answer these queries V(线段树-分类讨论)
题目链接:http://www.spoj.com/problems/GSS5/ 题意:给出一个数列.每次查询最大子段和Sum[i,j],其中i和j满足x1<=i<=y1,x2<=j& ...
- SPOJ 1557. Can you answer these queries II 线段树
Can you answer these queries II Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 https://www.spoj.com/pr ...
- bzoj 2482: [Spoj GSS2] Can you answer these queries II 线段树
2482: [Spoj1557] Can you answer these queries II Time Limit: 20 Sec Memory Limit: 128 MBSubmit: 145 ...
- spoj gss2 : Can you answer these queries II 离线&&线段树
1557. Can you answer these queries II Problem code: GSS2 Being a completist and a simplist, kid Yang ...
- SPOJ GSS1_Can you answer these queries I(线段树区间合并)
SPOJ GSS1_Can you answer these queries I(线段树区间合并) 标签(空格分隔): 线段树区间合并 题目链接 GSS1 - Can you answer these ...
随机推荐
- 如何获取wifi名称(SSID)
@import SystemConfiguration.CaptiveNetwork; /** Returns first non-empty SSID network info dictionary ...
- map--C++ STL 学习
map–C++ STL 学习 Map是STL的一个关联容器,它提供一对一(其中第一个可以称为关键字,每个关键字只能在map中出现一次,第二个可能称为该关键字的值)的数据处理能力. 说下map内 ...
- Razor 3、MVC 5
Razor 3 需要vs 2012 update 4 才可以 需要装一个 Microsoft ASP.NET and Web Tools 2013.1 才会有 MVC 5
- svn删除账户信息
当我们需要清理eclipse中记录的SVN账号信息时,按如下操作: eclipse中打开window------>preferences------->SVN页面,如下所示: 一般情况下, ...
- Eclipse 快速修复
Eclipse 快速修复 使用快速修复 在 Eclipse 编辑器中当你输入字母时,编辑器会对你输入的内容进行错误分析. Java 编辑器中使用 Java 语法来检测代码中的错误.当它发现错误或警告时 ...
- Oracle数据迁移expdp/impdp
Oracle数据迁移expdp/impdp目的:指导项目侧自行进行简单的数据泵迁移工作. 本文实验环境:Oracle 11.2.0.4,利用数据库自带的scott示例用户进行试验测试. 1.首先需要创 ...
- ps -ef | grep java 查看所有关于java的进程
ps -ef | grep java 查看所有关于java的进程
- (转)Unity 导出XML配置文件,动态加载场景
参考:http://www.xuanyusong.com/archives/1919 http://www.omuying.com/article/48.aspx 主要功能: 1.导出场景的配置文 ...
- sql 分组后查询最大所有列信息
CREATE TABLE students (course varchar(10), stu_name varchar(10), city varchar(10), score int ) inser ...
- 1.SpringMvc--初识springmvc
引自@精品唯居 springMvc是什么 springmvc是表现层的框架,是一个spring的表现层组件.是整个spring框架的一部分,但是也可以不使用springmvc.跟struts2框架功能 ...