看到这道题的第一个想法是二分+主席树(好暴力啊) 实际上不用这么麻烦,用一个双指针+桶扫一遍就行了 ~ code: #include <bits/stdc++.h> #define N 100006 #define setIO(s) freopen(s".in","r",stdin) using namespace std; int n,k,ans=1,kind,a[N],bu[N],A[N]; int main() { // setIO("i…

[bzoj 3048] [Usaco2013 Jan]Cow Lineup Description 给你一个长度为n(1<=n<=100,000)的自然数数列,其中每一个数都小于等于10亿,现在给你一个k,表示你最多可以删去k类数.数列中相同的数字被称为一类数.设该数列中满足所有的数字相等的连续子序列被叫做完美序列,你的任务就是通过删数使得该数列中的最长完美序列尽量长. Input Line 1: Two space-separated integers: N and K. Lines 2..…

3048: [Usaco2013 Jan]Cow Lineup Time Limit: 2 Sec  Memory Limit: 128 MBSubmit: 237  Solved: 168[Submit][Status][Discuss] Description Farmer John's N cows (1 <= N <= 100,000) are lined up in a row. Each cow is identified by an integer "breed ID&…

3048: [Usaco2013 Jan]Cow Lineup Time Limit: 2 Sec  Memory Limit: 128 MBSubmit: 225  Solved: 159[Submit][Status][Discuss] Description Farmer John's N cows (1 <= N <= 100,000) are lined up in a row. Each cow is identified by an integer "breed ID&…

BZOJ_3048_[Usaco2013 Jan]Cow Lineup _双指针 Description Farmer John's N cows (1 <= N <= 100,000) are lined up in a row. Each cow is identified by an integer "breed ID" in the range 0...1,000,000,000; the breed ID of the ith cow in the lineup…

一开始一脸懵逼.. 后来才想到维护一左一右俩指针l和r..表示[l,r]这段内不同种类的数字<=k+1种. 显然最左的.合法的l随着r的增加而不减. 顺便离散化,记一下各个种类数字出现的次数就可以算出答案了. 时间复杂度O(n) #include<cstdio> #include<iostream> #include<cstring> #include<algorithm> using namespace std; ; struct zs{int v,…

1636: [Usaco2007 Jan]Balanced Lineup Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 772  Solved: 560线段树裸题... Description For the daily milking, Farmer John's N cows (1 <= N <= 50,000) always line up in the same order. One day Farmer John decides to o…

1612: [Usaco2008 Jan]Cow Contest奶牛的比赛 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 891  Solved: 590[Submit][Status][Discuss] Description FJ的N(1 <= N <= 100)头奶牛们最近参加了场程序设计竞赛:).在赛场上,奶牛们按1..N依次编号.每头奶牛的编程能力不尽相同,并且没有哪两头奶牛的水平不相上下,也就是说,奶牛们的编程能力有明确的排名. 整个比…

RMQ.. ------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; ++i ) #define clr( x…

对于第 i 头牛 , 假如排名比它高和低的数位 n - 1 , 那么他的 rank 便可以确定 . floyd --------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep…