hqd_acm的专栏

线段树的题目#include #define LL(x) x << 1#define RR(x) x << 1 | 1const int MAXN = 200000 + 123;struct NODE{ int l, r, rmax, rmin; int lazy; int mid() { return (l + r) >> 1;

很不错的一道线段树的题目在每个节点里面加个线性表，这是关键的操作！线性表里面是这个节点的范围内，将它按照a数组从小到大排序，接着剔除掉a[i] >= a[j] 并且b[i] >= b[j]这种j另外还要用到(a[x] - a[y])*(b[z]-b[y

AC自动机+矩阵乘法+容斥原理刚开始是用单纯的状态压缩，当然是TLE后来用容斥原理，接着几乎快了64倍还多，接着各种弄，就过了2.99S，不知道那些弄到0.05秒的神牛怎么弄的

由二维的推一下就好了注意一下三维空间的容斥性，更新的时候注意一下这是区间更新，单点查询仔细想一想为什么可以这么做 Add(x1, y1, z1, 1); Add(x2+1, y1,

/*首先以1为根，扫描一遍树，得到每个节点的minchild[i][2]，儿子节点的最小值和次小值(不同子树的)和每个节点的最小后缀的值mindown[i](1)如果x是y的父节点的话，那么直接输出mindown[i],minchild[i][0]即可(2)如果y是x的父

小大中+小中大= ?#include #include const int MAXN = 100000 + 1234;const int mod = 100000007;int n;long long tree[MAXN];long long ans[MAXN

按右端点排序#include #include #include using namespace std;const int MAXN = 50010;const int MAXM = 200000 + 123;struct Edge{ int id,

首先以x为第一关键字，y为第二关键字进行排序，离散化得到一个长度为num的数组（任何一次查询对应的点都能在这个数组中找到）对此进行建树比如进行第i次操作找到其在数组中对应的点pos那么pos--->num这里面的所有点的x值都大于等于pos点看这一段序列是

操作十分不同，很特别比赛的时候没想出来，好好总结吧

去年杭州区域赛的一道题目，比较简单的AC自动机#include #include const int NODE = 1000;const int CH = 26;int tree[NODE][CH], cnt;int fail[NODE], word[NOD

#include #include #include using namespace std;#define LL(x) (x << 1)#define RR(x) (x << 1) | 1const int MAXN = 100000;struct Seg_T

前天比赛遇到一道简化版的抑或的线段树，比赛调了一个点才过这道题是上次的增强版#include #include #include using namespace std;#define LL(x) (x << 1)#define RR(x) (x <<

晚上发现NOI不给用优先级队列，就自己写个堆，还行，速度#include const int N = (500000 + 1234) * 2;struct heap_node{ int value, l, r, minv, loc; heap_

虽然不难，但是很讲究是否理解了并查集的实质#include #include const int MAXN = 10000 + 1234;int n, q;int con[MAXN];int p[MAXN];int num[MAXN];int find(

后缀数组变了个行就傻眼了。。。前面求出每个点为中心的最长回文字，接着对于一个[l,r]的查询，有两种二分方法第一是left = 1, right = r - l + 1, mid = (l+ r ) >>1, 对于[l+mid-1,r-mid+1]这个区间用RMQ求出一个最大值P

北邮邀请赛的题目，比赛的时候思路很乱，现在整理下吧。网上已经给出了有公式解法，我这里还是说说树状数组的解法吧题目只有两个操作，一个是将一个点的颜色取反，一个是询问有多少个端点颜色不同并且包含小于k的区间。总体思路是这样首先是用树状数组保存每个点的颜色，如果是1，那个点加1，这是初

#include #include const int MAXN = 1000000 + 123;struct node{ int id, val; node(int x, int y): id(x), val(y) {} node(){}}q[

#include #include #include using namespace std;const int MAXN = 100000 + 123;struct node{ int id, val; node(int x, int y): id(

#include #include const int MAXN = 100000 + 123;struct node{ int id, val; node(int x, int y): id(x), val(y) {} node() {}} q

Sliding WindowTime Limit: 12000MS Memory Limit: 65536KTotal Submissions: 18639 Accepted: 5331Case Time Limit: 5000MSDescriptionAn array of s

/*这题因为不是对数列第K个，而是对初始化标记的序号操作，所以标记的处理很不一般*/#include #include #include using namespace std;const int MAXN = 100000 + 123;const int INF

1470. Another Sequence ProblemProblem code: SEQ2You are to write a program to perform some operations on a given sequence.These operations a

#include const int MAXN = 100001;const int INF = 1 << 30;struct NODE{ int c[2], p, v, mul, sz; bool d;}T[MAXN];int n

#include #include const int MAXN = 100001;const int INF = 1 << 30 ;struct Node{ int c[2], p, v, mul, sz; bool d;}T[M

 Censored!Time Limit: 5000MS Memory Limit: 10000KTotal Submissions: 3608 Accepted: 999DescriptionThe alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word br

<br /> <br />考研路茫茫——单词情结Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)<br />Total Submission(s): 520    Accepted Submission(s): 143<br /><br /><br /><br />#include<cstdio>#include<cstring>typedef unsigned __int64 ul

DNA SequenceTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5080 Accepted: 1766DescriptionIt's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence，For example, if a ani

<br /> <br />病毒侵袭持续中Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)<br />Total Submission(s): 866    Accepted Submission(s): 293<br /><br /><br />Problem Description小t非常感谢大家帮忙解决了他的上一个问题。然而病毒侵袭持续中。在小t的不懈努力下，他发现了网路中的“万恶之源”

<br /> <br />病毒侵袭Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)<br />Total Submission(s): 1621    Accepted Submission(s): 423<br /><br /><br />Problem Description当太阳的光辉逐渐被月亮遮蔽，世界失去了光明，大地迎来最黑暗的时刻。。。。在这样的时刻，人们却异常兴奋——我们能在有

<br />Keywords SearchTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)<br />Total Submission(s): 7476    Accepted Submission(s): 2541<br /><br /><br />Problem DescriptionIn the modern time, Search engine came into the life

<br />Life FormsTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 4628 Accepted: 1207<br />Description<br />You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles

<br />Common SubstringsTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 3112 Accepted: 1002<br />Description<br />A substring of a string T is defined as:<br /> T(i, k)=TiTi+1...Ti+k-1, 1≤i≤i+k-1≤|T|.<br /> <br />Given two strings A, B and one i

<br />Maximum repetition substringTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2233 Accepted: 547<br />Description<br />The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same co

SPOJ Problem Set (classical)687. RepeatsProblem code: REPEATSA string s is called an (k,l)-repeat if s is obtained by concatenating k>=1 times some seed string t with length l>=1. For example, the strings = abaabaabaabais a (4,3)-repeat with t = aba as its

<br />1517. Freedom of ChoiceTime Limit: 2.0 second<br />Memory Limit: 32 MBBackgroundBefore Albanian people could bear with the freedom of speech (this story is fully described in the problem "Freedom of speech"), another freedom - the freedom of choice -

<br /> Power StringsTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 16624 Accepted: 6923<br />DescriptionGiven two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think o

<br /> <br />1297. PalindromeTime Limit: 1.0 second<br />Memory Limit: 16 MBThe “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S

<br /> Frequent valuesTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7310 Accepted: 2617<br />Description<br />You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries

<br /> Frequent valuesTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7308 Accepted: 2615<br />Description<br />You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries

<br />Balanced LineupTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 16743 Accepted: 7758Case Time Limit: 2000MS<br />Description<br />For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer J