解题报告

题意:

原本区间1到n都是1,区间成段改变成一个值,求最后区间1到n的和。

思路:

线段树成段更新,区间去和。

#include <iostream>
#include <cstdio>
#include <cstring> using namespace std;
int sum[500000],lz[500000];
void push_up(int root,int l,int r)
{
sum[root]=sum[root*2]+sum[root*2+1];
}
void push_down(int root,int l,int r)
{
if(lz[root]!=-1)
{
int m=(r-l+1);
lz[root*2]=lz[root];
lz[root*2+1]=lz[root];
sum[root*2]=lz[root]*(m-m/2);
sum[root*2+1]=lz[root]*(m/2);
lz[root]=-1;
}
}
void update(int root,int l,int r,int ql,int qr,int v)
{
if(ql>r||qr<l)return ;
if(ql<=l&&r<=qr)
{
lz[root]=v;
sum[root]=v*(r-l+1);
return ;
}
int mid=(l+r)/2;
push_down(root,l,r);
update(root*2,l,mid,ql,qr,v);
update(root*2+1,mid+1,r,ql,qr,v);
push_up(root,l,r);
}
int q_sum(int root,int l,int r ,int ql,int qr)
{
if(ql>r||qr<l)return 0;
if(ql<=l&&r<=qr)return sum[root];
int mid=(l+r)/2;
return q_sum(root*2,l,mid,ql,qr)+q_sum(root*2+1,mid+1,r,ql,qr);
}
int main()
{
int t,i,k,j,ql,qr,a,q,n;
while(~scanf("%d",&t))
{
k=1;
while(t--)
{
memset(lz,-1,sizeof(lz));
memset(sum,0,sizeof(sum));
scanf("%d%d",&n,&q);
for(i=1; i<=n; i++)
update(1,1,n,i,i,1);
for(i=1; i<=q; i++)
{
scanf("%d%d%d",&ql,&qr,&a);
update(1,1,n,ql,qr,a);
}
printf("Case %d: The total value of the hook is %d.\n",k++,q_sum(1,1,n,1,n));
}
}
return 0;
}

Just a Hook

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

Total Submission(s): 17182    Accepted Submission(s): 8575

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.

HDU1698_Just a Hook(线段树/成段更新)的更多相关文章

  1. HDU 1698 Just a Hook(线段树成段更新)

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

  2. HDU 1698 Just a Hook (线段树 成段更新 lazy-tag思想)

    题目链接 题意: n个挂钩,q次询问,每个挂钩可能的值为1 2 3,  初始值为1,每次询问 把从x到Y区间内的值改变为z.求最后的总的值. 分析:用val记录这一个区间的值,val == -1表示这 ...

  3. hdu698 Just a Hook 线段树-成段更新

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1698 很简单的一个线段树的题目,每次更新采用lazy思想,这里我采用了增加一个变量z,z不等于0时其绝 ...

  4. HDU-1698-Just a Hook-区间更新+线段树成段更新

    In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. T ...

  5. ACM: Copying Data 线段树-成段更新-解题报告

    Copying Data Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Description W ...

  6. Codeforces Round #149 (Div. 2) E. XOR on Segment (线段树成段更新+二进制)

    题目链接:http://codeforces.com/problemset/problem/242/E 给你n个数,m个操作,操作1是查询l到r之间的和,操作2是将l到r之间的每个数xor与x. 这题 ...

  7. POJ 2777 Count Color (线段树成段更新+二进制思维)

    题目链接:http://poj.org/problem?id=2777 题意是有L个单位长的画板,T种颜色,O个操作.画板初始化为颜色1.操作C讲l到r单位之间的颜色变为c,操作P查询l到r单位之间的 ...

  8. hdu 4747【线段树-成段更新】.cpp

    题意: 给出一个有n个数的数列,并定义mex(l, r)表示数列中第l个元素到第r个元素中第一个没有出现的最小非负整数. 求出这个数列中所有mex的值. 思路: 可以看出对于一个数列,mex(r, r ...

  9. HDU 3577 Fast Arrangement ( 线段树 成段更新 区间最值 区间最大覆盖次数 )

    线段树成段更新+区间最值. 注意某人的乘车区间是[a, b-1],因为他在b站就下车了. #include <cstdio> #include <cstring> #inclu ...

随机推荐

  1. Android 表格布局<TableLayout>

    表格布局即,tableLayout,表格布局通过行.列的形式来管理UI组件,TablelLayout并不需要明确地声明包含多少行.多少列,而是通过TableRow,以及其他组件来控制表格的行数和列数, ...

  2. 看来IT技术与军事技术都是相通的——都是对新事物极为敏感的领域

    这是读到这段时候的感想: 和海军中那些狂热的相信“皇军不可战胜”的大舰巨炮主义者们不同,山口对于与美国开战的主张是持坚定的反对态度的,和山本五十六都做过日本驻美武官的山口都认为一旦与美开战,日本或许能 ...

  3. java thread reuse(good)

    I have always read that creating threads is expensive. I also know that you cannot rerun a thread. I ...

  4. 用Python对体积较大的CSV文件进行比较的经验

    用Python对体积较大的CSV文件进行比较的经验 » 进化的测试 | 进化的测试 用Python对体积较大的CSV文件进行比较的经验  python  Add comments 八 032010   ...

  5. C++学习笔记10-面向对象

    1.  面向对象的程序设计是基于三个基本概念:数据抽象.继承和动态绑定. 在C++ 在,凭借一流的数据抽象,随着一类从一个类派生还继承:派生类的成员继承基类.决定是使用基类中定义的函数还是派生类中定义 ...

  6. POJ1505&amp;&amp;UVa714 Copying Books(DP)

    Copying Books Time Limit: 3000MS Memory Limit: 10000K Total Submissions: 7109 Accepted: 2221 Descrip ...

  7. Linux中利用crontab创建计划任务

    在linux中启动crontab服务: /etc/init.d/crond  start crontab的命令格式 crontab -l   显示当前的crontab 文件(默认编写的crontab文 ...

  8. Bean-Query 一个把对象转换为Map的Java工具库

    刚开源了一个经过完整測试的Java工具类. 地址例如以下: https://github.com/Jimmy-Shi/bean-query 使用说明例如以下: Bean-query Click Her ...

  9. 组件-------(一)redis系列--安装部署redis+实现redis分布式缓存 java+Spring+redis

    目的:解决单机session不能共享问题,插入查询数据库时间效率问题,实现分布式缓存. 准备材料:Redis 下载链接 http://pan.baidu.com/s/1dEGTxvV 相关jar包如果 ...

  10. 基于FP-Tree的关联规则FP-Growth推荐算法Java实现

    基于FP-Tree的关联规则FP-Growth推荐算法Java实现 package edu.test.ch8; import java.util.ArrayList; import java.util ...