Tallest Cow POJ - 3263 （区间点修改）

【Tallest Cow POJ - 3263 （区间点修改）】的更多相关文章

Tallest Cow POJ - 3263 （区间点修改）

FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of that cow…

POJ 3263 Tallest Cow 题解

题目 FJ's $N (1 ≤ N ≤ 10,000)$ cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height $H (1 ≤ H ≤ 1,000,000)$ of the tallest cow along with the index I…

poj 3263 Tallest Cow（线段树）

Language: Default Tallest Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 1964 Accepted: 906 Description FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which i…

洛谷P2879 [USACO07JAN]区间统计Tallest Cow

To 洛谷.2879 区间统计题目描述 FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with th…

bzoj1635 / P2879 [USACO07JAN]区间统计Tallest Cow

P2879 [USACO07JAN]区间统计Tallest Cow 差分对于每个限制$(l,r)$,我们建立一个差分数组$a[i]$ 使$a[l+1]--,a[r]++$,表示$(l,r)$区间内的数至少比$l,r$小$1$ 最后统计下前缀和,顺便把最大高度加上去. 记得判重. (逃一节晚自习真chiji) 不写了,放风时间到了. #include<iostream> #include<cstring> #include<cstdio> #include<alg…

[Luogu2879][USACO07JAN]区间统计Tallest Cow

题目描述 FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of tha…

poj3263 Tallest Cow

题意略去. 考虑给定的R对pair(A, B). 即A能看见B,这意味着B不比A低,并且区间内部的所有元素的高度严格小于A的高度. 我们规定区间的方向:若A > B,为反方向,反之称为正方向. 容易发现,区间在同一方向上不交叉,即要么相离,要么相互包含. 在相反反方向上,若两区间不相互包含,必定是有且仅有一个公共区间端点. 首先将所有idx上的高度默认设置为H. 对于相离的区间,他们之间的计算是独立的. 考虑相互包含的区间,先考虑范围最大的那个区间,处理之间闭区间[A,B]上的高度是相同的,满足…

POJ 3264 区间最大最小值Sparse_Table算法

题目链接:http://poj.org/problem?id=3264 Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 47515 Accepted: 22314 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up i…

poj 3264 区间最大最小值 RMQ问题之Sparse_Table算法

Balanced Lineup Time Limit: 5000 MS Memory Limit: 0 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main [Submit] [Status] [Discuss] Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order…

【BZOJ】1635: [Usaco2007 Jan]Tallest Cow 最高的牛（差分序列）

http://www.lydsy.com/JudgeOnline/problem.php?id=1635 差分序列是个好东西啊....很多地方都用了啊,,, 线性的进行区间操作orz 有题可知 h[a+1]~a[b-1]都是比h[a]和h[b]小,那么最佳方案就是将次区间的所有高度-1,那么我们就将整个区间-1 也就是sum[a+1]--, sum[b]++ 而条件h[a]>=h[b]我还不明觉厉啊..... (脑补:假设一般情况下h[a]==h[b]的,而却有c使得(a, c), pos[c]…

POJ3263 Tallest Cow 差分

题目描述 FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of tha…

BZOJ1635: [Usaco2007 Jan]Tallest Cow 最高的牛

1635: [Usaco2007 Jan]Tallest Cow 最高的牛 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 346 Solved: 184[Submit][Status] Description FJ's N (1 <= N <= 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height…

BZOJ 1635: [Usaco2007 Jan]Tallest Cow 最高的牛

题目 1635: [Usaco2007 Jan]Tallest Cow 最高的牛 Time Limit: 5 Sec Memory Limit: 64 MB Description FJ's N (1 <= N <= 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are t…

1635: [Usaco2007 Jan]Tallest Cow 最高的牛

1635: [Usaco2007 Jan]Tallest Cow 最高的牛 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 383 Solved: 211[Submit][Status] Description FJ's N (1 <= N <= 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height…

bzoj 1635: [Usaco2007 Jan]Tallest Cow 最高的牛——差分

Description FJ's N (1 <= N <= 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 <= H <= 1,000,000) of the tallest cow along with…

Tallest Cow

题目描述 FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of tha…

catch that cow POJ 3278 搜索

catch that cow POJ 3278 搜索题意原题链接 john想要抓到那只牛,John和牛的位置在数轴上表示为n和k,john有三种移动方式:1. 向前移动一个单位,2. 向后移动一个单位,3. 移动到当前位置的二倍处.输出移动的最少次数. 解题思路使用搜索,准确地说是广搜,要记得到达的位置要进行标记,还有就是减枝. 详情见代码实现. 代码实现 #include<cstdio> #include<cstring> #include<iostream>…

【差分】Tallest Cow

题目 FJ's N(1≤N≤10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H(1≤H≤1,000,000) of the tallest cow along with the index I of that cow. FJ ha…

【POJ 3263】 Tallest Cow

[题目链接] http://poj.org/problem?id=3263 [算法] 若A和B两头牛可以互相看见,那么说明中间的牛的高度都至少比它们少1,因此,我们可以引入一个差分数组c 对于每组关系,我们将c[a+1]减一,c[b]加一最后,每头牛的最高高度就是 h + c的前缀和 [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #includ…

poj 3263 Tallest Cow

一个压了很久的题目,确实很难想,看了别人的做法后总算明白了. 首先要明白一点,因为题目说明了不会有矛盾,所以题目给出来的区间是不能相交的,否则是矛盾的.(原因自己想) 然后既然区间只能是包含的,就很明显了,一个数被包含了几次它就要比最高的牛矮多少. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const i…

【差分】POJ 3263 Tallest Cow

题目大意 POJ链接给出$n$头牛的身高,和$m$对关系,表示牛$a[i]$与$b[i]$可以相互看见.已知最高的牛为第$p$头,身高为$h$. 求每头牛的身高最大可能是多少. (老姚的翻译好像改变量名了昂,但是无所谓) 输入格式第一行四个整数$n,p,h,m$,表示见题意, 接下来$m$行,每行两个正整数表示$a[i],b[i]$. 输出格式 $n$行整数表示每头牛的最大可能身高. 数据范围 \(1\le n\le 10000,1\le h\le 1…

POJ - 2528 区间离散化，线段树区间修改，区间询问

这个题非常有意思的地方是,我们发现区间[1,4]和[5,8]是紧挨着的,因为这个的数代表的是一段区间,原本我们对于普通的离散, a[1]=1,a[2]=5,a[3]=6,a[4]=8;数组下标就是重新离散的位置,但是a[2]和a[3]明显不重叠,为此我们需要重新考虑离散的内容,其实不妨这样,如果区间的间隔大于1,那么我们插入一个数a[i]+1,这样就强行把a[i]和a[i+1]分开,因为如三张海报为:1~10 1~4 6~10 离散化时 X[ 1 ] = 1, X[ 2 ] = 4, X[ 3…

洛谷 P2879 [USACO07JAN]区间统计Tallest Cow

传送门题目大意: n头牛,其中最高身高为h,给出r对关系(x,y) 表示x能看到y,当且仅当y>=x并且x和y中间的牛都比他们矮的时候,求每头牛的最高身高. 题解:贪心+差分将每头牛一开始都设为最高高度. 每一对关系(x,y),我们将[x+1,y-1]这个区间的身高变为 min(x,y)-1.这样是不对了.因为要维护[x+1,y-1]这个区间里各个元素的大小关系,所以要将[x+1,y-1]的元素身高都减1. 一开始我是用线段树做的,后来发现题解用的差分. 没有询问的区间修改,差分做就好了…

[USACO07JAN]区间统计Tallest Cow

前缀和 sum[i]表示前i个数的和每次读入a[i]的时候 sum[i] = sum[i - 1] + a[i]; 查询l ~ r区间的和: sum[r] - sum[l - 1] 差分即前缀和的修改操作我们定义pre[i]表示前i个数需要改变的值则对一个区间l ~ r + k的操作是 pre[r] + k; pre[l - 1] - k; 则多算的1 ~ l - 1部分就抵消,等价与l ~ r + k; 本题中同一关系可能给出多次,需用map或hash判重 #include<iostr…

题解 P2879 【[USACO07JAN]区间统计Tallest Cow】

题目链接: https://www.luogu.org/problemnew/show/P2879 思路: 先不管最大高度,我们读入一对x,y.说明,x+1~y-1之间牛的身高都小于x,y. 然后不妨将这个区间打个标记-1.所有操作后,可知最高的那个牛它的标记一定是0,并且标记数量与身高排名正相关,于是再将所有奶牛的标记数加上最高身高h就是可能的最大身高. 因为之前的操作保证标记的数量是按着身高排名来的,加上最大身高当然是最大可能身高. 然后我们是要将x+1~y-1打标记,根据差分数组的思维,就…