http://poj.org/problem?id=3468 (题目链接) 题意 给出一个序列,要求维护区间修改与区间求和操作. Solution 多年以前学习的树状数组区间修改又忘记了→_→. 其实就是用树状数组维护一个差分序列${delta[i]}$,${delta[x]}$记录${[i,n]}$中每一个数的增量,每次修改${[l,r]}$就转化为了${delta[l]+=d,delta[r+1]-=d}$. 对于求和操作${[l,r]}$,其实就是${sum(x)-sum(y)}$,我们这…

其实这两题都是基础的线段树,但对于我这个线段树的初学者来说,总结一下还是很有用的: poj3468显然是线段树区间求和,区间更改的问题,而poj2528是对区间染色,问有多少种颜色的问题: 线段树的建立和求和附代码,还是比较简单的: 这里想说的是区间修改,用到了了lazy思想:打标记: 拿poj2528举例,比如对区间[l,r]染色,我们只要在线段树中,被[l,r]覆盖的最大子区间[p,q]上标记被染成了什么颜色即可,不需要再往下遍历[p,q]的左右孩子:当下次修改影响到了区间[p,q]时(区间…

题目链接:poj3468 题意:给定一段数组,有两种操作,一种是给某段区间加c,另一种是查询一段区间的和 思路:暴力的方法是每次都给这段区间的点加c,查询也遍历一遍区间,复杂度是n*n,肯定过不去,另一种思路是用线段树记录区间的和,每次查询的复杂度是lgn,修改不必更新到每个点,当某个区间全被修改时,我们可以给它加一个懒惰标记,表示这个区间的所有下面节点都需要更新,只是因为现在不需要使用而暂时没有更新.这样修改的复杂度也降到了lgn ac代码: #include <iostream> #inc…

https://vjudge.net/problem/POJ-3468 线段树区间更新(lazy数组)模板题 #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #include<map> #define lson l, m, rt<<1 #define…

poj3468 A Simple Problem with Integers 题意:O(-1) 思路:O(-1) 线段树功能:update:成段增减 query:区间求和 Sample Input 10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4 Sample Output 4 55 9 15 Hint The sums may exceed the range of 32-bit integers. 题目大意:   N个数 M…

题目链接:https://vjudge.net/problem/POJ-3468 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 nu…

洛谷P3372 //线段树 询问区间和,支持区间修改 #include <cstdio> using namespace std; struct treetype { int l,r; long long sum,d; }; ; treetype a[maxn<<]; long long num[maxn]; void build(int k,int l,int r) { a[k].l=l;a[k].r=r;a[k].d=; if (a[k].l==a[k].r) { a[k].s…

A Simple Problem With Integers POJ-3468 这题是区间更新的模板题,也只是区间更新和区间查询和的简单使用. 代码中需要注意的点我都已经标注出来了,容易搞混的就是update函数里面还需要计算sum数组.因为这里查询的时候是直接用sum查询结点. //区间更新,区间查询 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #inc…

此题是一道线段树的裸题,这里只是为了保存我的zkw线段树模板 #include <cstdio> #include <cstring> #include <iostream> using namespace std; typedef long long LL; inline int geti() { static int Ina; static char Inc; static bool InSign; InSign = false; while ((Inc = get…

题目链接:http://poj.org/problem?id=3468 这题是线段树的题,拿来学习treap. 不旋转的treap. #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <map> #include <set> #include <string> #include <bitset>…

Description 给出了一个序列,你需要处理如下两种询问. "C abc"表示给[a, b]区间中的值全部增加c (-10000 ≤ c ≤ 10000). "Q ab" 询问[a, b]区间中所有值的和. Input 第一行包含两个整数N, Q.1≤ N,Q ≤ 100000. 第二行包含n个整数,表示初始的序列A (-1000000000 ≤ Ai ≤1000000000). 接下来Q行询问,格式如题目描述. Output 对于每一个Q开头的询问,你需要输…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 97196   Accepted: 30348 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 92127   Accepted: 28671 Case Time Limit: 2000MS 描述 You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operatio…

用splay做了一遍. 建树时是按照数列序号从小到大排好的,每个节点左子树的序号小于右子树的序号及这个节点本身.由于查询[l,r]要伸展l-1,r+1所以我们要多加2个结点,保证边界处理时不出问题.由于这样每次查找l-1时,要找的应该是l(r+1也是找r+2). #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #define ll __int64…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 60745   Accepted: 18522 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

题目地址http://poj.org/problem?id=3468 题目大意很简单,有两个操作,一个 Q a, b 查询区间[a, b]的和 C a, b, c让区间[a, b] 的每一个数+c 第一次线段树的延时标记,花了好大的功夫才写好==! 很容易看出来使用使用线段树记录区间的和,但是难点在于每次修改的是一个区间而不是一个点 所以采用的方法就是每次做修改操作时,只将区间［a,b］的标记+c,而不是真正意义上的将区间[a, b] 的每一个值+c. 而当我们做查询操作时,就只需要将区间[a,…

今天学了很多关于树状数组的技巧.一个是利用树状数组可以简单的实现段更新,点询问(二维的段更新点询问也可以),每次修改只需要修改2个角或者4个角就可以了,另外一个技巧就是这题,原本用线段树做,现在可以用树状数组做的题,只需多维护一个bit即可.具体的思路见下面的链接: http://hi.baidu.com/billdu/item/053f6a15ca301b0a8ebde400 要理解里面的橙色块求的时候是打竖看的,不是打横看的. #pragma warning(disable:4996) #i…

#include<cstdio> int lb,rb,data; long long sum[5000000],extra[5000000]; void add(int l,int r,int now) {     if(lb<=l&&rb>=r){         extra[now]+=data;         sum[now]+=data*(r-l+1);         return;     }     int mid=(l+r)/2,nl=2*now,…

题目链接:http://poj.org/problem?id=3468 以前用线段树做过,现在用Splay Tree A了,向HH.kuangbin.cxlove大牛学习了各种Splay各种操作,,,Orz.. Splay Tree的区间操作和线段树的操作差不多,也是保存子树的值,然后懒惰操作,在Rotate()最后维护节点信息的时候,只要Push_Up(y)的,因为x还需要网上旋转到根节点,最后更新下就可以了,并且在下一次Rotate()的时候,还会Push_Down(x)的信息,因此不能Pu…

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…

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…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 92921   Accepted: 28910 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers Time Limit:5000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3468 Appoint description:  System Crawler  (2014-11-12) Description You have N integers, A1, A2, ... , AN. You nee…

题目大意: 2个操作 A.区间a b 增加 c B 查询a b; 注意事项:1.记住要清除标记 2.查询时要下放标记,但没必要向上更新 线段:自带的,不用建模 区间和性质:sum: /* WA 1次 以为不要LONG LONG */ #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <ctime> #include <algor…

转载请注明出处:http://blog.csdn.net/u012860063 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 th…

题目连接:http://poj.org/problem?id=3468 线段树功能:update:成段增减 query:区间求和. 分析:需要用到延迟标记(或者说懒惰标记),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or询问到的时候. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define LL long long #…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 57666   Accepted: 17546 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 47174   Accepted: 13844 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers 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…

我的线段树真的没救了......还是多练几道吧....... 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…