HDU 1698 Just a Hook(线段树】的更多相关文章

Just a Hook Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1698 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 cons…
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[1] 就是answer [时间复杂度]\(O(nlogn)\) &代码: #include <bits/stdc++.h> using namespace std; const int maxn = 100000 + 9 ; int n,q,x,y,z; int seg[maxn<&…
Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 23229    Accepted Submission(s): 11634 Problem Description In the game of DotA, Pudge’s meat hook is actually the most horrible thing…
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表示这个区间值不统一,而且已经向下更新了, val != -1表示这个区间值统一, 更新某个区间的时候只需要把这个区间分为几个区间更新就行了, 也就是只更新到需要更新的区间,不用向下更新每一个一直到底了,在更新的过程中如果遇到之前没有向下更新的, 就需要向下更新了,因为这个区间的值已经不统一了. 其实这就…
Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18378    Accepted Submission(s): 9213 Problem Description In the game of DotA, Pudge’s meat hook is actually the most horrible thing f…
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…
Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 30553    Accepted Submission(s): 15071 Problem Description In the game of DotA, Pudge's meat hook is actually the most horrible thing…
描述 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 n…
题目地址:pid=1698">HDU 1698 区间替换裸题.相同利用lazy延迟标记数组,这里仅仅是当lazy下放的时候把以下的lazy也所有改成lazy就好了. 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h> #include <math.h> #includ…
来谈谈自己对延迟标记(lazy标记)的理解吧. lazy标记的主要作用是尽可能的降低时间复杂度. 这样说吧. 如果你不用lazy标记,那么你对于一个区间更新的话是要对其所有的子区间都更新一次,但如果用lazy标记的话. 就只需要更新这一个区间然后加一个标记,那么如果要访问这个区间的子区间,因为有lazy标记,所以下次访问会将区间的lazy标记传递给子区间,让后去更新子区间,这样我们不必在每次区间更新操作的时候更新该区间的全部子区间,等下次查询到这个区间的时候只需要传递lazy标记就可以了 但从时…