https://www.luogu.org/problem/show?pid=3029 题目描述 Farmer John has hired a professional photographer to take a picture of some of his cows. Since FJ's cows represent a variety of different breeds, he would like the photo to contain at least one cow fro…

思路 考虑比较朴素的解法,枚举每个长度为\(k+1\)的区间,然后统计区间中出现次数最多的颜色.这样的话复杂度为\(O(n*k)\)的,显然不行. 观察到统计每个区间中出现次数最多的颜色中,可以只用看每种颜色在区间中出现的最后一个位置,这样的话只需要我们开个桶统计一下数量就行. 所以就类似于尺取那样,维护颜色种类不超过\(k+1\)的区间,对于每次新加进来的值令其\(cnt++\),并且维护ans.当颜色种类超过\(k+1\)时,就减去区间前面的值,因为它们与后面的颜色不可能再连在一起了. 因为…

P3080 [USACO13MAR]牛跑The Cow Run 题目描述 Farmer John has forgotten to repair a hole in the fence on his farm, and his N cows (1 <= N <= 1,000) have escaped and gone on a rampage! Each minute a cow is outside the fence, she causes one dollar worth of dam…

P2853 [USACO06DEC]牛的野餐Cow Picnic 题目描述 The cows are having a picnic! Each of Farmer John's K (1 ≤ K ≤ 100) cows is grazing in one of N (1 ≤ N ≤ 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 ≤ M ≤ 10,000) one-way path…

P2853 [USACO06DEC]牛的野餐Cow Picnic 题目描述 The cows are having a picnic! Each of Farmer John's K (1 ≤ K ≤ 100) cows is grazing in one of N (1 ≤ N ≤ 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 ≤ M ≤ 10,000) one-way path…

题目描述 The cows are having a picnic! Each of Farmer John's K (1 ≤ K ≤ 100) cows is grazing in one of N (1 ≤ N ≤ 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 ≤ M ≤ 10,000) one-way paths (no path connects a pasture to…

题目描述 Farmer John has hired a professional photographer to take a picture of some of his cows. Since FJ's cows represent a variety of different breeds, he would like the photo to contain at least one cow from each distinct breed present in his herd. F…

题目描述 Like everyone else, FJ is always thinking up ways to increase his revenue. To this end, he has set up a series of tolls that the cows will pay when they traverse the cowpaths throughout the farm. The cows move from any of the N (1 <= N <= 250)…

传送门 题目大意: m个车道. 如果第i头牛前面有k头牛,那么这头牛的最大速度会 变为原本的速度-k*D,如果速度小于l这头牛就不能行驶. 题解:贪心 让初始速度小的牛在前面 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #define N 50009 using namespace std; int n,m,d,l,k,ans; int a[N],…

原题链接 假设只有一个政党,那么这题就退化成求树的直径的问题了,所以我们可以从此联想至\(k\)个政党的情况. 先处理出每个政党的最大深度,然后枚举每个政党的其它点,通过\(LCA\)计算长度取\(\max\)即可. 因为枚举只是枚举该政党的所有点,所以总的枚举复杂度依旧是\(O(n)\),总复杂度\(O(nlog_2n)\). #include<cstdio> #include<cmath> using namespace std; const int N = 2e5 + 10;…