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.

注意后面这个标点‘.’

代码：

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<map>

#include<set>

#include<vector>

#include<cmath>

const int maxn=1e5+5;

typedef long long ll;

using namespace std;

struct node

{

	ll l,r,sum;

}tree[maxn<<2];

ll lazy[maxn<<2];

void pushup(int m)

{

	tree[m].sum=tree[m<<1].sum+tree[m<<1|1].sum;

}

void pushdown(int m,int l)

{

	if(lazy[m]!=0)

	{

		lazy[m<<1]=lazy[m];

		lazy[m<<1|1]=lazy[m];

		tree[m<<1].sum=lazy[m]*(l-(l>>1));

		tree[m<<1|1].sum=lazy[m]*(l>>1);

		lazy[m]=0;

	}

}

void build(int m,int l,int r)

{

	tree[m].l=l;

	tree[m].r=r;

	lazy[m]=0;

	if(l==r)

	{

	    tree[m].sum=1;

		return;

	}

	int mid=(l+r)>>1;

	build(m<<1,l,mid);

	build(m<<1|1,mid+1,r);

	pushup(m);

}

void update(int m,int l,int r,int val)

{

	if(tree[m].l==l&&tree[m].r==r)

	{

		lazy[m]=val;

		tree[m].sum=(ll)val*(r-l+1);

		return;

	}

	if(tree[m].l==tree[m].r)

	return;

	int mid=(tree[m].l+tree[m].r)>>1;

	pushdown(m,tree[m].r-tree[m].l+1);

	if(r<=mid)

	{

		update(m<<1,l,r,val);

	}

	else if(l>mid)

	{

		update(m<<1|1,l,r,val);

	}

	else

	{

		update(m<<1,l,mid,val);

		update(m<<1|1,mid+1,r,val);

	}

	pushup(m);

}

ll query(int m,int l,int r)

{

	if(tree[m].l==l&&tree[m].r==r)

	{

		return tree[m].sum;

	}

	pushdown(m,tree[m].r-tree[m].l+1);

	int mid=(tree[m].l+tree[m].r)>>1;

	ll res=0;

	if(r<=mid)

	{

		res+=query(m<<1,l,r);

	}

	else if(l>mid)

	{

		res+=query(m<<1|1,l,r);

	}

	else

	{

		res+=(query(m<<1,l,mid)+query(m<<1|1,mid+1,r));

	}

	return res;

}

int main()

{

    int T;

    cin>>T;

    int n,m;

    int cnt=1;

    while(T--)

    {

	  cin>>n;

	  build(1,1,n);

	  cin>>m;

	  char op[2];

	  int l,r,val;

	  for(int t=0;t<m;t++)

	  {

	  	scanf("%d%d%d",&l,&r,&val);

	  	update(1,l,r,val);

	  }

	  printf("Case %d: The total value of the hook is %lld.\n",cnt,query(1,1,n));

	  cnt++;

	}

	return 0;

}

Just a Hook(线段树区间修改值)-------------蓝桥备战系列的更多相关文章

  1. 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 ...

  2. 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[ ...

  3. (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 ...

  4. 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 ...

  5. 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 ...

  6. [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 ...

  7. 【原创】hdu1698 Just a Hook(线段树→区间更新，区间查询)

    学习线段树第二天,这道题属于第二简单的线段树,第一简单是单点更新,这个属于区间更新. 区间更新就是lazy思想,我来按照自己浅薄的理解谈谈lazy思想: 就是在数据结构中,树形结构可以线性存储(线性表 ...

  8. hdu - 1689 Just a Hook (线段树区间更新)

    http://acm.hdu.edu.cn/showproblem.php?pid=1698 n个数初始每个数的价值为1,接下来有m个更新,每次x,y,z 把x,y区间的数的价值更新为z(1<= ...

  9. hdu1698 Just a hook 线段树区间更新

    题解: 和hdu1166敌兵布阵不同的是 这道题需要区间更新(成段更新). 单点更新不用说了比较简单,区间更新的话,如果每次都更新到底的话,有点费时间. 这里就体现了线段树的另一个重要思想:延迟标记. ...

随机推荐

  1. mybatis内部类映射写法

    直接上代码:   mybatis内部类映射使用$而不是.

  2. 766. Toeplitz Matrix斜对角矩阵

    ［抄题］: A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same element. Now ...

  3. opennebula模板对照比较

    良好模板 有问题模板

  4. CentOS7下源码包方式安装rabbitmq

    1.先安装erlang http://www.cnblogs.com/justphp/p/6093880.html 2.下载rabbitmq rpm包: wget http://www.rabbitm ...

  5. App测试从入门到精通之兼容性和回归测试

    兼容性测试需要考虑的APP测试点 1.不同网络环境下的兼容性测试 2.不同手机操作系统兼容性测试 3.不同应用软件的兼容性测试 4.不同的容量大小的SIM卡之间的互相兼容测试 5.当安装杀毒软件时,应 ...

  6. TensorFlow安装教程

    Windows7 安装TensorFlow(本人试了好多方法后的成果):https://www.cnblogs.com/bxyan/p/6869237.html Linux: sudo pip ins ...

  7. Head First Python之人人都爱列表（1-初识Python）

    IDLE 内置函数==BIF==built-in function 默认地,内置函数都是紫色,字符串是绿色,关键字是橙色. tab:自动补全 Alt-P:前一个 Alt-N:下一个 列表的特性 列表看 ...

  8. ArcGIS Desktop和Engine中对点要素图层Graduated Symbols渲染的实现 Rotation Symbol （转）

    摘要         ArcGIS中,对于要素图层的渲染,支持按照要素字段的值渲染要素的大小,其中Graduated Symbols可以对大小进行分级渲染.在个人开发系统的过程中,也可以用来美化数据显 ...

  9. oracle 中 创建序列sequence

    drop sequence SEQ_YCXWP_CGD; create sequence SEQ_YCXWP_CGD increment start nomaxvalue;

  10. HackFifteen 移除背景以提升Activity启动速度

    1.概要:     移除窗口默认背景是提升应用程序启动速度的一个简单技巧.判断是否需要移除背景的原则很简单:如果应用程序界面需要 占据窗口100%的控件,就需要将windowBackground属性设 ...