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 y 表示把第x种操作一直到第y种操作都执行一遍 最终输出在k个操作结束之后的序列 题目大意 就是线段树的成段更新嘛~~~先用线段树统计每种操作的次数,然后再执行m次成段更新,最后查询到底的查询即可~~~树状数组也可搞,似乎写起来还更简单些~~~还有一个更犀利的O(n)的算法,不过我暂时还没弄懂~~…
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…
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 mos…
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<&…
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…
转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1698 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 consecut…
线段树裸题.自己写复杂了,准确说是没想清楚就敲了. 先是建点为已插花之和,其实和未插花是一个道理,可是开始是小绕,后来滚雪球了,跪了. 重新建图,分解询问1为:找出真正插画的开始点和终止点,做成段更新. 再次向notonlysuccess大神致谢,清晰的代码+清晰的思路=ac #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define lson l,m,rt&…
HDU  1698 链接:  http://acm.hdu.edu.cn/showproblem.php?pid=1698 线段树功能:update:成段替换 (由于只query一次总区间,所以可以直接输出1结点的信息) <span style="font-size:18px;">#include<iostream> #include<cstdio> #include<cstring> #define lson l,m,rt<<…
Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 33311 Accepted: 10058 Description Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem. The…
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 53169 Accepted: 15897 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of ope…
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…
Mayor's posters Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 37346Accepted: 10864 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at…
A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 59046   Accepted: 17974 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…
hdu1556 Color the ball Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9143    Accepted Submission(s): 4677 Problem Description N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的"小飞…
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 . 建树的时候初始化为1,更新区间时候放懒惰标记,下推标记更新区间和. 由于是替换,不是累加,所以更新的时候不是+=,而是直接=. 注意这点就可以了,然后就是多组数据注意memset,因为这个WA几发. 代码总览 #include <bits/stdc++.h> #define maxn 200010 #defin…
描述 很多学校流行一种比较的习惯.老师们很喜欢询问,从某某到某某当中,分数最高的是多少. 这让很多学生很反感.不管你喜不喜欢,现在需要你做的是,就是按照老师的要求,写一个程序,模拟老师的询问.当然,老师有时候需要更新某位同学的成绩. Input 本题目包含多组测试,请处理到文件结束.在每个测试的第一行,有两个正整数 N 和 M ( 0<N<=200000,0<M<5000 ),分别代表学生的数目和操作的数目.学生ID编号分别从1编到N.第二行包含N个整数,代表这N个学生的初始成绩,…
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<=z<=3),问更新完后的总价值. 线段树的区间更新,需要用到延迟标记,简单来说就是每次更新的时候不要更新到底,用延迟标记使得更新延迟到下次需要更新或者询问的时候. 这题只需要输出总区间的信息,即直接输出1结点的信息. #include <iostream> #include <cstd…
题意:给你个n,表示区间[1,n],价值初始为1,给你m段区间和价值,更新区间,区间价值以最后更新为准,问更新后区间价值总和为多少 思路:两种方法,可以先存下来,倒过来更新,一更新节点马上跳出,比较快 线段树 #include <iostream> using namespace std; int data[100005][3]; int main() { int t,q,n,i,j,sum,k,v; scanf("%d",&t); for(i=1;i<=t;…
一条钩子由许多小钩子组成 更新一段小钩子 变成铜银金 价值分别变成1 2 3 输出最后的总价值 Sample Input11021 5 25 9 3 Sample OutputCase 1: The total value of the hook is 24. # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmat…
线段树的基本知识可以先google一下,不是很难理解 线段树功能:update:单点增减 query:区间求和 #include <bits/stdc++.h> #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 using namespace std; const int MAXN = 50008; int sum[MAXN<<2]; void build(int l, int r, int rt)…
Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. In…
hdu1698 Just a Hook 题意:O(-1) 思路:O(-1) 线段树功能:update:成段替换 (由于只query一次总区间,所以可以直接输出1结点的信息) 题意:给一组棍子染色,不同的颜色有不同的值,执行一系列的区间染色后,问这组棍子的总值是多少. #include <cstdio> #include <algorithm> using namespace std; #define lson l , m , rt << 1 #define rson m…
Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15889    Accepted Submission(s): 7897 Problem Description In the game of DotA, Pudge's meat hook is actually the most horrible thing…
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 numb…