hdu 1698 Just a Hook(线段树之 成段更新)
Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit:
32768/32768 K (Java/Others)
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.
1
10
2
1 5 2
5 9 3
Case 1: The total value of the hook is 24.
#include<cstdio> #define lson l, mid, root<<1
#define rson mid+1, r, root<<1|1
const int N = 100010;
struct node
{
int l;
int r;
int sum;
int color;
}a[N<<2]; void PushUp(int root)
{
a[root].sum = a[root<<1].sum + a[root<<1|1].sum;
}
void PushDown(int len, int root)
{
if(a[root].color)
{
a[root<<1].color = a[root<<1|1].color = a[root].color;
a[root<<1].sum = (len - (len>>1)) * a[root].color;
a[root<<1|1].sum = (len>>1) * a[root].color;
a[root].color = 0;
}
}
void build_tree(int l, int r, int root)
{
a[root].l = l;
a[root].r = r;
a[root].color = 0; if(l == r)
{
a[root].sum = 1;
return ;
}
int mid = (l + r) >> 1;
build_tree(lson);
build_tree(rson);
PushUp(root);
}
void update(int l, int r, int root, int z)
{
if(l <= a[root].l && r >= a[root].r)
{
a[root].color = z;
a[root].sum = (a[root].r - a[root].l + 1) * z;
return;
}
PushDown(a[root].r - a[root].l + 1, root);
int mid = (a[root].l + a[root].r) >> 1;
if(l <= mid) update(l, r, root<<1, z);
if(r > mid) update(l, r, root<<1|1, z);
PushUp(root);
} int main()
{
int T, n, m, cas = 0;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
build_tree(1, n, 1);
scanf("%d",&m);
int x, y, z;
while(m--)
{
scanf("%d%d%d",&x,&y,&z);
update(x, y, 1, z);
}
int ans = a[1].sum;
printf("Case %d: The total value of the hook is %d.\n", ++cas, ans);
}
return 0;
}
hdu 1698 Just a Hook(线段树之 成段更新)的更多相关文章
- Codeforces295A - Greg and Array(线段树的成段更新)
题目大意 给定一个序列a[1],a[2]--a[n] 接下来给出m种操作,每种操作是以下形式的: l r d 表示把区间[l,r]内的每一个数都加上一个值d 之后有k个操作,每个操作是以下形式的: x ...
- 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 ...
- HDU 1698 just a hook 线段树,区间定值,求和
Just a Hook Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1 ...
- 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 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 ...
- [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 ...
- (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+lazy-target 区间刷新
Just a Hook Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tota ...
- HDU 1698 Just a Hook(线段树区间更新查询)
描述 In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes ...
随机推荐
- HDU 4893 Wow! Such Sequence!(2014年多校联合 第三场 G)(线段树)
磨了一天的线段树,不能说完全搞清楚,只能说有一个大概的了解,靠着模板才把这道题A了,只能说太弱~~! 题意: 初始时有一字符串,全为0. 三种操作: 1 k d - add 把d加到第k个数上去2 ...
- 党建凯,创新工场知乎团队Web前端工程师
Nicholas C. Zakas谈怎样才能成为优秀的前端工程师: 昨天,我负责了Yahoo!公司组织的一次面试活动,感触颇深的是其中的应聘者提问环节.我得说自己对应聘者们提出的大多数问题都相当失望. ...
- c++中编译器的作用
编译器的部分工作是寻找程序代码中的错误.编译器不能查出程序的意义是否正确. 但它能够查出程序形式上的错误.以下是编译器能查出的最普遍的一些错误: (1)语法错误.程序猿犯了c++语言中的语法错误. ( ...
- 在StatusBar中显示当前时间
在StatusBar中显示当前时间,如下: 1.在String Table中插入一项 (注意:状态栏将根据字符串的长度来确定相应窗格的缺省宽度,所以指定为00:00:00就为时间的显示预留了空间) ...
- Xtrabackup使用指南 | 简单.生活
Xtrabackup使用指南 | 简单.生活 Xtrabackup是一个对InnoDB做数据备份的工具,支持在线热备份(备份时不影响数据读写),是商业备份工具InnoDB Hotbackup的一个很好 ...
- Python 新浪微博元素 (Word, Screen Name)词汇多样性
CODE: #!/usr/bin/python # -*- coding: utf-8 -*- ''' Created on 2014-7-10 @author: guaguastd @name: w ...
- Mybatis在oracle、mysql、db2、sql server的like模糊查询
<!-- oracle --> <select id="searchUserBySearchName" parameterType="java.lang ...
- Android获取设备採用的时间制式(12小时制式或24小时制式)
/** * 获取设备採用的时间制式(12小时制式或者24小时制式) * 注意: * 在模拟器上获取的时间制式为空 */ private void getTime_12_24(Context conte ...
- codeforces 598D Igor In the Museum
题目链接:http://codeforces.com/problemset/problem/598/D 题目分类:dfs 题目分析:处理的时候一次处理一片而不是一个,不然会超时 代码: #includ ...
- TControl的消息覆盖函数大全(15个WM_函数和17个CM_函数,它的WndProc就处理鼠标与键盘消息)
注意,这些函数只有Private一种形式(也就是不允许覆盖,但仍在动态表格中)(特别注意,这里居然没有WM_PAINT函数): TControl = class(TComponent) private ...