#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define R(a,b,c) for(register int a = (b); a <= (c); ++ a) #define nR(a,b,c) for(register int a = (b); a >= (c); -- a) #defi…

题目描述 Farmer John's new barn consists of a huge circle of N stalls (2 <= N <= 3,000,000), numbered 0..N-1, with stall N-1 being adjacent to stall 0. At the end of each day, FJ's cows arrive back at the barn one by one, each with a preferred stall the…

题面 当然可以用并查集做,不过你需要按秩合并+路径压缩(才可能过),因为数据范围十分不友好...... USACO的官方做法更为优秀.首先题目告诉我们牛们加入的前后顺序不影响结果(自己证明也很容易,显然两头牛到达一个房子最后的结果看起来是一样的).所以我们不妨先把牛们都安排在它们喜欢的房子,然后$O(n)$扫描一遍模拟把它们放进对应的房子这个过程.因为第一次我们先扫到前面,可能后面一些牛之后又到了前面,所以要再扫一遍. #include<cstdio> #include<cstring&…

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…

Problem 1. Empty Stalls 扫两遍即可. Problem 2. Line of Sight 我们发现能互相看见的一对点一定能同时看见粮仓的某一段.于是转换成有n段线段,问有多少对线段相交.可以按左端点排序,用优先队列维护右端点,弹出比左端点小的. 为了方便计算对数,我们可以先做一遍,再把每个线段都+2*pi,再计数. Problem 3. No Change (没有看到要买的东西必须是依次的..) 如果要依次买的话就显然可以用dp搞.…

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…

动态规划属于技巧性比较强的题目,如果看到过原题的话,对解题很有帮助 55. Jump Game Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Determine if you ar…

唯一路径问题II Unique Paths II Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is…

The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his wa…

题意:给定一个数N,表示有N个位置,要么放置0,要么放置1,问至少存在一个连续的M个1的放置方式有多少? 分析:正面求解可能还要考虑到重复计算带来的影响,该题适应反面求解.设dp[i][j]表示到前 i 为后导 1 个数为 j 的方案数,于是有动态规划方程: dp[i][0] = sum{ dp[i-1][0... min(i-1, M) ] };dp[i][j] = dp[i-1][j-1]  其中 j != 1 单单根据这个方程时间度为O(N*M),还是不足以在有限的时间内解出该问题.通过观…