1698-Just a Hook 线段树(区间替换)
Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32141 Accepted Submission(s):
15804
most horrible thing for most of the heroes. The hook is made up of several
consecutive metallic sticks which are of the same length.
Now Pudge wants to
do some operations on the hook.
Let us number the consecutive metallic
sticks of the hook from 1 to N. For each operation, Pudge can change the
consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver
sticks or golden sticks.
The total value of the hook is calculated as the sum
of values of N metallic sticks. More precisely, the value for each kind of stick
is calculated as follows:
For each cupreous stick, the value is 1.
For
each silver stick, the value is 2.
For each golden stick, the value is
3.
Pudge wants to know the total value of the hook after performing the
operations.
You may consider the original hook is made up of cupreous
sticks.
line of the input is the number of the cases. There are no more than 10
cases.
For each case, the first line contains an integer N,
1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and
the second line contains an integer Q, 0<=Q<=100,000, which is the number
of the operations.
Next Q lines, each line contains three integers X, Y,
1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the
sticks numbered from X to Y into the metal kind Z, where Z=1 represents the
cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden
kind.
the total value of the hook after the operations. Use the format in the
example.
#include<cstdio>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int MAXN = ;
int sum[MAXN<<];
int col[MAXN<<];
int t,n,q;
void putup(int rt)
{
sum[rt] = sum[rt<<]+sum[rt<<|];
}
void putdown(int rt,int m)
{
if (col[rt])
{
col[rt<<] = col[rt<<|] = col[rt];
sum[rt<<] = (m-(m>>))*col[rt];
sum[rt<<|] = (m>>)*col[rt];
col[rt] = ;
}
}
void build(int l,int r,int rt)
{
col[rt] = ;
sum[rt] = ;
if (l==r) return ;
int m = (l+r)>>;
build(lson);
build(rson);
putup(rt);
}
void update(int l,int r,int rt,int L,int R,int c)
{
if (L<=l && r<=R)
{
col[rt] = c;
sum[rt] = c*(r-l+);
return ;
}
putdown(rt,r-l+);
int m = (l+r)>>;
if (L<=m) update(lson,L,R,c);
if (R>m) update(rson,L,R,c);
putup(rt);
}
int main()
{
scanf("%d",&t);
for (int i=; i<=t; ++i)
{
scanf("%d%d",&n,&q);
build(,n,);
while (q--)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
update(,n,,x,y,z);
}
printf("Case %d: The total value of the hook is %d.\n",i,sum[]);
}
return ;
}
1698-Just a Hook 线段树(区间替换)的更多相关文章
- HDU 1698 Just a Hook(线段树 区间替换)
Just a Hook [题目链接]Just a Hook [题目类型]线段树 区间替换 &题解: 线段树 区间替换 和区间求和 模板题 只不过不需要查询 题里只问了全部区间的和,所以seg[ ...
- [HDU] 1698 Just a Hook [线段树区间替换]
Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total ...
- HDU 1698 Just a Hook(线段树区间替换)
题目地址:pid=1698">HDU 1698 区间替换裸题.相同利用lazy延迟标记数组,这里仅仅是当lazy下放的时候把以下的lazy也所有改成lazy就好了. 代码例如以下: # ...
- HDU.1689 Just a Hook (线段树 区间替换 区间总和)
HDU.1689 Just a Hook (线段树 区间替换 区间总和) 题意分析 一开始叶子节点均为1,操作为将[L,R]区间全部替换成C,求总区间[1,N]和 线段树维护区间和 . 建树的时候初始 ...
- (简单) HDU 1698 Just a Hook , 线段树+区间更新。
Description: In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of ...
- HDU 1698 Just a Hook(线段树区间更新查询)
描述 In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes ...
- HDU 1698 Just a Hook 线段树区间更新、
来谈谈自己对延迟标记(lazy标记)的理解吧. lazy标记的主要作用是尽可能的降低时间复杂度. 这样说吧. 如果你不用lazy标记,那么你对于一个区间更新的话是要对其所有的子区间都更新一次,但如果用 ...
- hdu1698(线段树区间替换模板)
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1698 题意: 第一行输入 t 表 t 组测试数据, 对于每组测试数据, 第一行输入一个 n , 表示 ...
- poj2528(线段树区间替换&离散化)
题目链接: http://poj.org/problem?id=2528 题意: 第一行输入一个 t 表 t 组输入, 对于每组输入: 第一行 n 表接下来有 n 行形如 l, r 的输入, 表在区 ...
- Just a Hook 线段树 区间更新
Just a Hook In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of t ...
随机推荐
- 如何在SAP CRM里创建和消费Web service
Created by Wang, Jerry, last modified on Dec 19, 2014 The following steps demonstrates how to expose ...
- Jerry的WebClient UI 42篇原创文章合集
我要感谢CRM On Premise, 因为在这个产品上做开发让我得以使用WebClient UI框架.有些朋友觉得这个SAP自己发明的基于HTML+ABAP的MVC框架,和现在流行的三驾马车(Ang ...
- 2018.12.15 struts.xml 一般配置文件写法 && 配置动态方法
struts.xml 原始配置文件 配置 <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE s ...
- 使用Vue做评论+localStorage存储(js模块化)
未分模块化 html <!DOCTYPE html> <html> <head> <meta charset="utf-8"> &l ...
- P2852 [USACO06DEC]牛奶模式Milk Patterns
link 这是一道后缀匹配的模板题 我们只需要将height算出来 然后二分一下答案就可以了 #include<cstdio> #include<algorithm> #inc ...
- Android学习笔记_18_Activity生命周期 及 跳转方式
一.Activity有三个状态: 1.当它在屏幕前台时(位于当前任务堆栈的顶部),它是激活或运行状态.它就是响应用户操作的Activity. 2. 当它上面有另外一个Activity,使它失去了焦点但 ...
- Android学习笔记_15_网络通信之文件断点下载
一.断点下载原理: 使用多线程下载文件可以更快完成文件的下载,多线程下载文件之所以快,是因为其抢占的服务器资源多.如:假设服务器同时最多服务100个用户,在服务器中一条线程对应一个用户,100条线程在 ...
- asp.net mvc Post上传文件大小限制 (转载)
最近发现在项目中使用jQuery.form插件上传比较大的文件时,上传不了,于是改了下web.config的上传文件最大限制. <configuration> <system.web ...
- 用$(this)选择其下带有class的子元素
$(this).find('.son').removeClass("disn")
- vue webpack多页面构建
项目示例地址: https://github.com/ccyinghua/webpack-multipage 项目运行: 下载项目之后 # 下载依赖 npm install # 运行 npm run ...