Fiber Communications 总时间限制:  1000ms 内存限制:  65536kB 描述 Farmer John wants to connect his N (1 <= N <= 1,000) barns (numbered 1..N) with a new fiber-optic network. However, the barns are located in a circle around the edge of a large pond, so he can on…

Fiber Communications Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3804   Accepted: 1160 Description Farmer John wants to connect his N (1 <= N <= 1,000) barns (numbered 1..N) with a new fiber-optic network. However, the barns are loc…

1550: Fiber Communications  Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByteTotal Submit: 3            Accepted:2 Description Farmer John wants to connect his N (1 <= N <= 1,000) barns (numbered 1..N) with a new fiber-optic network…

Fiber Communications Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 4236   Accepted: 1276 Description Farmer John wants to connect his N (1 <= N <= 1,000) barns (numbered 1..N) with a new fiber-optic network. However, the barns are loc…

Description Farmer John wants to connect his N (1 <= N <= 1,000) barns (numbered 1..N) with a new fiber-optic network. However, the barns are located in a circle around the edge of a large pond, so he can only connect pairs of adjacent barns. The ci…

第二次参加USACO 本来打算2016-2017全勤的 January的好像忘记打了 听群里有人讨论才想起来铂金组三题很有意思,都是两个排列的交叉对问题 我最后得分889/1000(真的菜) T1.Why Did the Cow Cross the Road题目大意:给出两个N个排列(N<=100,000),允许把其中一个排列循环移动任意位,a[i]表示i在第一个排列中的位置,b[i]表示第二个,定义交叉对(i,j)满足a[i]<a[j]且b[i]>b[j],求最少交叉对.思路:数字大小…

1.Circular Barn   http://www.usaco.org/index.php?page=viewproblem2&cpid=621 贪心 #include <cstdio> #include <vector> #include <algorithm> #include <cstring> using namespace std; long long sum(long long v) { )*(*v+)/; } int main()…

原题地址:http://poj.org/problem?id=1944 题目大意:有n个点排成一圈,可以连接任意两个相邻的点,给出 p 对点,要求这 p 对点必须直接或间接相连,求最少的连接边数 数据范围:n <= 1000, p <= 10000 算法分析: 一开始当最小生成树做的,才发现自己 SB 了…… 先考虑不是环形而是线形的结构,直接贪心连接每两个点之间的所有点就好了.这样我们可以枚举环形的断点,然后逐次贪心,求最小解即可 很多同学在贪心的时候应用了线段树是复杂度高达O(np log…

Description Farmer John's family pitches in with the chores during milking, doing all the chores as quickly as possible. At FJ's house, some chores cannot be started until others have been completed, e.g., it is impossible to wash the cows until they…

那天打cf前无聊练手 T1.Why Did the Cow Cross the Road 题目大意:N*N的矩阵,从左上角走到右下角,走一步消耗T,每走3步消耗当前所在位置上的权值,求最小消耗 思路:好像很傻逼但我不会做,写BFS写着写着写成SPFA,管他呢,过了就行(大致就是每个点拆成三个,就能知道什么时候要加上额外的权值) #include<cstdio> #include<algorithm> using namespace std; <<],*S=B,C;int…