贪心问题,其实我觉得贪心就是合理的考虑最优情况,证明贪心可行即可.这题目没话多久一次ac.这道题需要注意房间号的奇偶性.1 3.2 4的测试数据.答案应该为20. #include <stdio.h> #include <stdlib.h> #define MAXNUM 505 ]; int visit[MAXNUM]; int comp(const void *a, const void *b) { return *(int *)a - *(int *)b; } int main…

Moving Tables Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24316    Accepted Submission(s): 8053 Problem Description The famous ACM (Advanced Computer Maker) Company has rented a floor of a b…

Moving Tables Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14418    Accepted Submission(s): 4939 Problem Description The famous ACM (Advanced Computer Maker) Company has rented a floor of a b…

Moving Tables Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 20344    Accepted Submission(s): 6900 Problem Description The famous ACM (Advanced Computer Maker) Company has rented a floor of a…

Moving Tables Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 20302    Accepted Submission(s): 6889 Problem Description The famous ACM (Advanced Computer Maker) Company has rented a floor of a…

Problem Description The famous ACM (Advanced Computer Maker) Company has rented a floor of a building whose shape is in the following figure. The floor has 200 rooms each on the north side and south side along the corridor. Recently the Company made…

Problem Description The famous ACM (Advanced Computer Maker) Company has rented a floor of a building whose shape is in the following figure. The floor has 200 rooms each on the north side and south side along the corridor. Recently the Company made…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…

http://www.lydsy.com/JudgeOnline/problem.php?id=1050 表示被暴力吓到了orz 我竟然想不到...我竟然还想到分数规划,,但是不可做...然后又想到最小生成树,,然后不会做orz 我一直在纠结怎么最大化(或最小化)分母和最小化(或最大化)分子的做法.....但是....不会orz 没想到是暴力orz 直接排序后枚举最小的边,生成树后要最大的边最小(排序后即可orz),然后更新答案即可. 但是不知道之前写错了啥wa了两发,,,随便改改才过orz #…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1050 这道题目隔了很久才做出来的.一开始把判断走廊有重叠的算法都想错了.以为重叠只要满足,下一次moving的起始room小于或等于上一次moving的结束room则证明有重复.这样只能保证局部不能同时进行moving,但是根本得不出其他moving哪些是可以同时进行搬动的. 正确的思路是,统计最大的重叠数,再乘以10即可.具体做法:把每个房间之间的走廊作为一个统计单位,当所有的办公桌都搬运完成之后…