题面 大意:让你把两个n的排列做匹配,连线不想交,而且匹配的数字的差<=4,求最大匹配数 sol:(参考了kczno1的题解)对于第一个排列从左往右枚举,用树状数组维护到达另一个序列第i个数字的最大值. #include <cstdio> #include <algorithm> using namespace std; ; int n,a[N],b[N],f[N],t[N]; #define lowbit(x) ((x)&(-x)) inline void ins(…

P3662 [USACO17FEB]Why Did the Cow Cross the Road II S 题目描述 The long road through Farmer John's farm has NN crosswalks across it, conveniently numbered 1 \ldots N1…N (1 \leq N \leq 100,0001≤N≤100,000). To allow cows to cross at these crosswalks, FJ in…

P3663 [USACO17FEB]Why Did the Cow Cross the Road III S 题目描述 Why did the cow cross the road? Well, one reason is that Farmer John's farm simply has a lot of roads, making it impossible for his cows to travel around without crossing many of them. 为什么牛过…

//神题目(题目一开始就理解错了)... 题目描述 Why did the cow cross the road? Well, one reason is that Farmer John's farm simply has a lot of roads, making it impossible for his cows to travel around without crossing many of them. 奶牛们为什么要穿马路?一个原因只是因为FJ的牧场的路实在是太多了,使得奶牛们每…

题目背景 给定长度为2N的序列,1~N各处现过2次,i第一次出现位置记为ai,第二次记为bi,求满足ai<aj<bi<bj的对数 题目描述 The layout of Farmer John's farm is quite peculiar, with a large circular road running around the perimeter of the main field on which his cows graze during the day. Every morn…

题目地址 又是一道奶牛题 从左到右扫描,树状数组维护[左端点出现而右端点未出现]的数字的个数.记录每个数字第一次出现的位置. 若是第二次出现,那么删除第一次的影响. #include <cstdio> #include <cstring> #define re register #define GC getchar() #define Lowbit(X) (X&(-X)) #define Clean(X,K) memset(X,K,sizeof(X)) int Qread…

本题是练习前缀和的好题!我们可以枚举前端点,确定一个长度为k的区间,然后利用前缀和统计区间内损坏的灯的数量,最后取最小值即可.AC代码: #include <bits/stdc++.h> using namespace std; inline int read()//快速读入 { ,x=; char c=getchar(); ') { ; c=getchar(); } ') { x=x*+c-'; c=getchar(); } return f*x; } ],s[],sum=; int mai…

题目描述 The long road through Farmer John's farm has  crosswalks across it, conveniently numbered  (). To allow cows to cross at these crosswalks, FJ installs electric crossing signals, which light up with a green cow icon when it is ok for the cow to c…

嘟嘟嘟 考虑dp. 对于ai,和他能匹配的bj只有9个,所以我们考虑从这9个状态转移. 对于ai 能匹配的一个bj,当前最大的匹配数一定是[1, j - 1]中的最大匹配数 + 1.然后用树状数组维护前缀匹配数最大值就行了. #include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<cstring> #include<cstdlib>…

[题目链接] https://www.lydsy.com/JudgeOnline/problem.php?id=4990 [算法] 首先记录b中每个数的出现位置 , 记为P 对于每个ai , 枚举(ai - 4) - (ai + 4) , 将Pj从大到小加入序列 然后求最长上升子序列即可 , 详见代码 时间复杂度 : O(NlogN) [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 1000010 int n , l…