Just a Hook(区间set)
Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29106 Accepted Submission(s): 14396
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.
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.
10
2
1 5 2
5 9 3
#include<bits/stdc++.h>
#define N 100005
#define lson i*2,l,m
#define rson i*2+1,m+1,r
#define ll long long
using namespace std;
long long sum[N*];
long long setv[N*];
void pushup(int i)
{
sum[i]=sum[i*]+sum[i*+];
} void pushdown(int i,int num)
{
if(setv[i])
{
sum[i*]=setv[i]*(num-num/);
sum[i*+]=setv[i]*(num/);
setv[i*]=setv[i];
setv[i*+]=setv[i];
setv[i]=;
}
} void build(int i,int l,int r)
{
setv[i]=;
sum[i]=;
if(l==r)
return ;
int m=(l+r)/;
build(lson);
build(rson);
pushup(i);
} void update(int ql,int qr,int val,int i,int l,int r)
{
if(ql<=l&&r<=qr)
{
setv[i]=val;
sum[i]=val*(r-l+);
return ;
}
pushdown(i,r-l+);
int m=(l+r)/;
if(m>=ql) update(ql,qr,val,lson);
if(m<qr) update(ql,qr,val,rson);
pushup(i);
} ll query(int ql,int qr,int i,int l,int r)
{
//cout<<"l="<<l<<" r="<<r<<endl;
if(ql<=l&&r<=qr)
{
return sum[i];
}
pushdown(i,r-l+);
int m=(l+r)/;
ll cur=;
if(m>=ql) cur+=query(ql,qr,lson);
if(m<qr) cur+=query(ql,qr,rson);
//cout<<cur<<endl;
return cur;
} int t,n,q;
int main()
{
//freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
scanf("%d",&t);
for(int Case=;Case<=t;Case++)
{
scanf("%d",&n);
build(,,n);
//for(int i=1;i<18;i++)
// cout<<sum[i]<<" ";
//cout<<endl;
int a,b,c;
scanf("%d",&q);
while(q--)
{
scanf("%d%d%d",&a,&b,&c);
//cout<<a<<" "<<b<<" "<<c<<endl;
update(a,b,c,,,n);
}
printf("Case %d: The total value of the hook is %lld.\n",Case,query(,n,,,n));
}
return ;
}
Just a Hook(区间set)的更多相关文章
- HDU1698 Just a Hook (区间更新)
Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total ...
- HDU 1698 Just a Hook 区间更新 lazy标记
lazy标记 #include <iostream> #include <cstdio> #include <cstring> #include <sstre ...
- 线段树入门&lazy思想
线段树将区间分成若干个子区间,子区间又继续分,直到区间为一个点(区间左值等于右值) 对于父区间[a,b],其子区间为[a,(a+b)/2]和[(a+b)/2+1,b] 用于求区间的值,如区间最值.区间 ...
- 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/区间更新)
题目链接: 传送门 Minimum Inversion Number Time Limit: 1000MS Memory Limit: 32768 K Description In the g ...
- HDU1698Just a Hook(线段树 + 区间修改 + 求和)
题目链接 分析:1-N区间内初始都是1,然后q个询问,每个询问修改区间[a,b]的值为2或3或者1,统计最后整个区间的和 本来想刷刷手速,结果还是写了一个小时,第一个超时,因为输出的时候去每个区间查找 ...
- hdu 1698:Just a Hook(线段树,区间更新)
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)
Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tota ...
- HDU 1698 Just a Hook (线段树区间更新)
题目链接 题意 : 一个有n段长的金属棍,开始都涂上铜,分段涂成别的,金的值是3,银的值是2,铜的值是1,然后问你最后这n段总共的值是多少. 思路 : 线段树的区间更新.可以理解为线段树成段更新的模板 ...
- HDU 1698 (线段树 区间更新) Just a Hook
有m个操作,每个操作 X Y Z是将区间[X, Y]中的所有的数全部变为Z,最后询问整个区间所有数之和是多少. 区间更新有一个懒惰标记,set[o] = v,表示这个区间所有的数都是v,只有这个区间被 ...
随机推荐
- 初识Hibernate之环境搭建
相信所有做后端的程序员同行们,没有不知道Hibernate大名的.这是一个经典的轻量级Java EE持久层的解决方案,它使得我们程序员能以面向对象的思维操作传统的关系型数据库,这也是其存在的 ...
- Thread.Join 和 Task.Wait 方法
这两个方法 可以说是类似的功能,都是对当前任务进行等待阻塞,执行完毕后再进行后续处理 talk is cheap, show you code,下面一个是异步执行,一个是加了阻塞,可以对比不同执行结果 ...
- vue学习之vue基本功能初探
vue学习之vue基本功能初探: 采用简洁的模板语法将声明式的将数据渲染进 DOM: <div id="app"> {{ message }} </div> ...
- Undefined symbols for architecture arm64: "_OBJC_CLASS_$_WKWebView", referenced from: objc-c
出现: Undefined symbols for architecture arm64: "_OBJC_CLASS_$_WKWebView", referenced from: ...
- Linux入门之常用命令(13) date
在linux shell编程中,经常用到日期的加减运算 以前都是自己通过expr函数计算,很麻烦 其实date命令本身提供了日期的加减运算 非常方便.例如:得到昨天的时间date +%Y%m%d -- ...
- Sum It Up 广搜
Sum It Up Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit St ...
- hdu4632
Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/ ...
- 关于xamarin.forms 中 list 的loadmore
前言 最近几天在研究上拉加载啊,下拉刷新啊什么的.然而坑爹的事情总是那么多.在xamarin.forms中,list自带的,并没有上拉加载的这个属性(难道当初他们封装方法时,就不会想到数据多了会咋整吗 ...
- IDEA Maven 三层架构 2、运行 springMVC
运行 SpringMVC 首先要理解 SpringMVC 应用程序的入口是配置文件 web.xml,其路径为"src/main/webapp/WEB-INF/web.xml",通过 ...
- cocos2dx - 环境配置,项目创建
准备工具 cocos2dx当前最新版本:v3.9 官网下载地址: http://www.cocos.com/download/ python 2.7x最新版本:2.7.11 官 ...