首先将排列和整个序列以及询问都反过来,问题变成给定一个位置$x$,问它经过若干轮置换后会到达哪个位置. 每次置换之后窗口都会往右滑动一个,因此其实真实置换是$p[i]-1$. 对于每个询问,求出轮数,倍增找到最终位置,注意当中途走到$0$时,说明离开了窗口,应及时终止. 时间复杂度$O((m+q)\log n)$. #include<cstdio> const int N=100010,M=30; int n,m,q,i,j,x,r,k,a[M][N]; inline void read(in…

[BZOJ4094][Usaco2013 Dec]Optimal Milking Description Farmer John最近购买了N(1 <= N <= 40000)台挤奶机,编号为1 ... N,并排成一行.第i台挤奶机每天能够挤M(i )单位的牛奶 (1 < =M(i) <=100,000).由于机器间距离太近,使得两台相邻的机器不能在同一天使用.Farmer Jo hn可以自由选择不同的机器集合在不同的日子进行挤奶.在D(1 < = D < = 50,00…

4094: [Usaco2013 Dec]Optimal Milking Description Farmer John最近购买了N(1 <= N <= 40000)台挤奶机,编号为1 ... N,并排成一行.第i台挤奶机每天能够挤M(i)单位的牛奶 (1 < =M(i) <=100,000). 由于机器间距离太近,使得两台相邻的机器不能在同一天使用.Farmer John可以自由选择不同的机器集合在不同的日子进行挤奶. 在D(1 < = D < = 50,000)天中…

4097: [Usaco2013 dec]Vacation Planning Description Air Bovinia is planning to connect the N farms (1 <= N <= 200) that the cows live on. As with any airline, K of these farms (1 <= K <= 100, K <= N) have been selected as hubs. The farms are…

Description Air Bovinia is planning to connect the N farms (1 <= N <= 200) that the cows live on. As with any airline, K of these farms (1 <= K <= 100, K <= N) have been selected as hubs. The farms are conveniently numbered 1..N, with farms…

Description Farmer John has N cows that need to be milked (1 <= N <= 10,000), each of which takes only one unit of time to milk. Being impatient animals, some cows will refuse to be milked if Farmer John waits too long to milk them. More specificall…

[题目链接] https://www.lydsy.com/JudgeOnline/problem.php?id=4093 [算法] 对于k个枢纽 , 分别在正向图和反向图上跑dijkstra最短路 , 即可 时间复杂度 : O(K(N + M) log N) [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 20010 #define MAXK 210 typedef long long LL; const LL in…

Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…

bzoj上的usaco题目还是很好的(我被虐的很惨. 有必要总结整理一下. 1592: [Usaco2008 Feb]Making the Grade 路面修整 一开始没有想到离散化.然后离散化之后就很好做了.F[I,j]表示第i个点,高度>=j或<=j,f[I,j]=min(f[i-1,j]+abs(b[j]-a[i]),f[I,j-1]) 1593: [Usaco2008 Feb]Hotel 旅馆 线段树 ★1594: [Usaco2008 Jan]猜数游戏 二分答案然后写线段树维护 15…

听说KPM初二暑假就补完了啊%%% 先刷Gold再刷Silver(因为目测没那么多时间刷Silver,方便以后TJ2333(雾 按AC数降序刷 ------------------------------------------------------------------------------------------------------- bzoj1597: [Usaco2008 Mar]土地购买  斜率优化DP h升序,w降序. f[i]=min(f[j]+h[i]*w[j+1])…