Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description
In the game of DotA, Pudge’s meat hook is actually the 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.
 
Input
The input consists of several test cases. The first 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.
 
Output
For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.
 
Sample Input
1
10
2
1 5 2
5 9 3
 
Sample Output
Case 1: The total value of the hook is 24.
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <algorithm>
#define LL long long using namespace std; const int Max = 110000; typedef struct Node
{
int val;//记录区间的价值 int data;//记录区间的共同的值(区间的值不相同的为零)
}Tree; Tree Tr[Max*4]; int N,Q; void Push_up(int st)//统计区间的信息
{
Tr[st].val = Tr[st<<1].val + Tr[st<<1|1].val; if(Tr[st<<1].data==Tr[st<<1|1].data&&Tr[st<<1].data)
{
Tr[st].data=Tr[st<<1].data;
}
else
{
Tr[st].data=0;
}
} void Push_down(int st,int L,int R)
{
if(Tr[st].data)
{
Tr[st<<1].data = Tr[st<<1|1].data = Tr[st].data; int mid = (L+R)>>1; Tr[st<<1].val=(mid-L+1)*Tr[st<<1].data; Tr[st<<1|1].val=(R-mid)*Tr[st<<1|1].data;
}
} void Build(int L,int R,int st)
{
Tr[st].data = 0; Tr[st].val = 0; if(R == L)
{
Tr[st].data = 1; Tr[st].val = 1; return ;
} int mid = (L+R)>>1; Build(L,mid,st<<1); Build(mid+1,R,st<<1|1); Push_up(st);
} void Update(int L,int R, int st, int l, int r, int d)
{
if(L > r ||R < l)
{
return ;
}
if(L>=l&&R<=r)
{
Tr[st].data = d; Tr[st].val=d*(R-L+1); return ;
} Push_down(st,L,R); int mid = (L+R)>>1; if(l<=mid)
{
Update(L,mid,st<<1,l,r,d);
}
if(r>mid)
{
Update(mid+1,R,st<<1|1,l,r,d);
} Push_up(st);
} int main()
{
int T; int x,y,d; scanf("%d",&T); for(int z = 1; z <= T;z++)
{
scanf("%d %d",&N,&Q); Build(1,N,1); while(Q--)
{
scanf("%d %d %d",&x,&y,&d); Update(1,N,1,x,y,d);
} printf("Case %d: The total value of the hook is %d.\n",z,Tr[1].val);
}
return 0;
}

 

Just a Hook(HDU1698 线段树的简单应用)的更多相关文章

  1. hdu1698 线段树区间更新

    Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Tota ...

  2. 『zkw线段树及其简单运用』

    阅读本文前,请确保已经阅读并理解了如下两篇文章: 『线段树 Segment Tree』 『线段树简单运用』 引入 这是一种由\(THU-zkw\)大佬发明的数据结构,本质上是经典的线段树区间划分思想, ...

  3. HDU1698 Just a Hook 【线段树】+【成段更新】+【lazy标记】

    Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Tota ...

  4. 【HDU1698】 Just a Hook 【线段树入门】

    原题:原题链接 题意:(机器翻译的...) 让我们将钩子的连续金属棒从1到N编号.对于每次操作,Pudge可以将连续的金属棒(从X到Y编号)改为铜棒,银棒或金棒. 钩的总值计算为N个金属棒的值的总和. ...

  5. hdoj 1698 Just a Hook 【线段树 区间更新】

    题目大意:有一段链子.初始的时候是铜的(价值为1),n代表有n段(1~n),输入a, b, c三个数分别表示将从a到b的链子的价值改为c, 最后问你经过多次改变之后的总价值. 策略:这道题是简单的线段 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. hdu 1698 Just a Hook(线段树之 成段更新)

    Just a Hook                                                                             Time Limit: ...

随机推荐

  1. 使用RMAN备份数据库

    1.1使用控制文件备份全库 1.1.1配置备份路径 1.1.2 RMAN备份全库 1.2使用catalog数据库备份全库 1.2.1配置到远程数据库的TNS 1.2.2创建表空间及设置用户 1.2.3 ...

  2. 配置DNS实验一例

    1安装bind软件 2查看当前DNS服务 3修改配置文件 4测试

  3. 2016HUAS暑假集训训练题 G - Oil Deposits

    Description The GeoSurvComp geologic survey company is responsible for detecting underground oil dep ...

  4. 修改RectTransform的宽度和高度

    rectTransform.sizeDelta = new Vector2( width, height);

  5. Web前端开发基础 第四课(CSS文字和段落排版)

    文字排版--字体 我们可以使用css样式为网页中的文字设置字体.字号.颜色等样式属性.下面我们来看一个例子,下面代码实现:为网页中的文字设置字体为宋体. body{font-family:" ...

  6. Windows Server 2003修改远程桌面最大连接数的方法

    Windows Server 2003远程桌面允许的终端连接数默认情况下只有2个线程,也就是2个用户,当我们需要多人同时登录服务器的时候,就需要适当增加远程连接同时在线的用户数. 修改远程桌面最大连接 ...

  7. IOS第九天(3:QQ聊天界面通知的使用)

    #import <Foundation/Foundation.h> #import "Person.h" #import "XQCompany.h" ...

  8. JavaBean组件的基本使用-语法

    <jsp:useBean id="实例化对象名称" scope="保存范围" class="包.类名"> </jsp:us ...

  9. DS实验题 Order 已知父节点和中序遍历求前、后序

    题目: 思路: 这题是比较典型的树的遍历问题,思路就是将中序遍历作为位置的判断依据,假设有个节点A和它的父亲Afa,那么如果A和Afa的顺序在中序遍历中是先A后Afa,则A是Afa的左儿子,否则是右儿 ...

  10. controlling the variance of request response times and not just worrying about maximizing queries per second

    http://highscalability.com/blog/2010/11/4/facebook-at-13-million-queries-per-second-recommends-minim ...