P2859 [USACO06FEB]摊位预订Stall Reservations 题目描述 Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B.…

P2859 [USACO06FEB]摊位预订Stall Reservations 维护一个按右端点从小到大的优先队列 蓝后把数据按左端点从小到大排序,顺序枚举. 每次把比右端点比枚举线段左端点小的数据从优先队列中删掉. 在整个过程中队列的最大长度即为答案. 总之用优先队列模拟一下就ok了 对于luogu需要输出方案数的问题: 再开一个优先队列存未用的编号 每次有线段进队时取走最小的编号,出队时再还回来. 似乎暴力也行(大雾) #include<iostream> #include<cst…

[USACO06FEB]摊位预订Stall Reservations 题目描述 Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviou…

题目链接: https://www.luogu.org/problemnew/show/P2859 思路: 首先大家会想到这是典型的贪心,类似区间覆盖问题的思路,我们要将每段时间的左端点从小到大排序,然后一个个插入,插入时比较是否先前的牛棚中已经有牛挤完了奶,如果没有就新增一个牛棚,否则用挤完奶的牛棚. 如果插入时扫描一遍找可用的牛棚的话肯定是会超时的,那么我们就要用一个priority_queue(当然你手写堆也可以)维护先前插入的牛棚中最早挤完奶,即右端点最小的那个. Tips: 注意排序后…

贪心 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; const int MAXN = 50005; int init() { int rv = 0, fh = 1; char c = getchar(); while(c < '0' || c > '9…

P2858 [USACO06FEB]奶牛零食Treats for the Cows 题目描述 FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period ti…

题目描述 FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time. The treats are interesting for many re…

https://www.luogu.org/problem/show?pid=1118#sub 题目描述 FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. Th…

#include<iostream> using namespace std ; ; int y[N][N]; int n; int a[N]; bool st[N]; int sum; bool flag; void print() { ; i<=n; i++) cout<<a[i]<<" "; } void dfs(int step,int ans) { if(ans>sum||flag) return ; &&ans…

[USACO06FEB] Stall Reservations 贪心 \(n\)头牛,每头牛占用时间区间\([l_i,r_i]\),一个牛棚每个时间点只能被一头牛占用,问最少新建多少个牛棚,并且每头牛在哪个牛棚里? 比较巧的\(O(n)\)扫一遍做法,用一个小跟堆维护所有牛棚最后一个牛占用的时间(即\(r_i\)),贪心的想,如果最先结束的那个牛棚都不能满足当前牛,那么我们只能新开一个牛棚,并继续维护小根堆:如果那个牛棚满足,那么这个牛就去那个牛棚,更新当前牛棚最后占用时间. #include…