Tallest Cows(差分) n头牛,给出最高牛的位置和身高,其他牛身高未知,给出m对相对关系,表示可以相互看见当且仅当他们中间的牛都比他们矮.求每头牛身高最大值是多少. 差分数组的性质:前缀和为原序列 给出相对关系可以考虑使用差分,a,b可以看见,则将a,b之间的所有的牛身高c[]减1,使用差分数组d[]维护.最后差分数组的前缀和就是和最高牛的相对关系(最高牛c[]的值肯定为0,他不可能在任何可以相互看见的两头牛中间) AC Code: #include <bits/stdc++.h>…

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…

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…

开个坑记录一下刷USACO的Training的记录 可能会随时弃坑 只有代码和做法简述 可能没有做法简述 [USACO1.1]你的飞碟在这儿Your Ride Is He… 模拟,细节已忘 #include<iostream> #include<cstdio> #include<cstring> using namespace std; ],s2[]; int main() { ,ss2=; scanf("%s",&s1); scanf(&q…

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…

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…

s[i]为差分后的"i这头牛前有几头比它高",计算答案的时候加成前缀和,假设第一头最高减一下即可 用map记录一下被加过的区间,避免重复 #include<iostream> #include<cstdio> #include<map> #include<algorithm> using namespace std; const int N=10005; int n,q,h,r,s[N],f[N],a,b; map<int,bool…

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…

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]…

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…