HDU 2795 Billboard(宣传栏贴公告,线段树应用) ACM 题目地址:HDU 2795 Billboard 题意:  要在h*w宣传栏上贴公告,每条公告的高度都是为1的,并且每条公告都要尽量贴最上面最靠左边的,给你一系列的公告的长度,问它们能不能贴上. 分析:  不是非常好想,只是想到了就非常好写了.  仅仅要把宣传栏倒过来就好办了,这时候就是变成有h条位置能够填公告,填放公告时就能够尽量找最左边的合适的位置来放了.  能够用线段树实现,查找的复杂度是O(logn),须要注意的坑点…
Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15129    Accepted Submission(s): 7506 Problem Description In the game of DotA, Pudge’s meat hook is actually the most horrible thing f…
Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1444    Accepted Submission(s): 329 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…
Multiply game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3224    Accepted Submission(s): 1173 Problem Description Tired of playing computer games, alpc23 is planning to play a game on numbe…
HDU 1394 Minimum Inversion Number(线段树求最小逆序数对) ACM 题目地址:HDU 1394 Minimum Inversion Number 题意:  给一个序列由[1,N]构成.能够通过旋转把第一个移动到最后一个.  问旋转后最小的逆序数对. 分析:  注意,序列是由[1,N]构成的,我们模拟下旋转,总的逆序数对会有规律的变化.  求出初始的逆序数对再循环一遍即可了. 至于求逆序数对,我曾经用归并排序解过这道题:点这里.  只是因为数据范围是5000.所以全…
任意门:http://acm.hdu.edu.cn/showproblem.php?pid=2795 Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 28743    Accepted Submission(s): 11651 Problem Description At the entrance to the un…
Billboard Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2795 Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2795 题目大意:有一块h*w的矩形广告板,要往上面贴广告;   然后给n个1*wi的广告,要求把广告贴上去: 而且要求广告要尽量往上贴并且尽量靠左;  求每个广告的所在的位置,不能贴则为-1. 用线段树模拟,要是左子树的最大值比当前广告大,就查询更新左子树,否则就右子树. #include <iostream> #include <cstdio> #include <cstrin…
题目大意:有一个h*w的公告榜,可以依次在上面添加信息.每个信息的长度为x,高为1. 优先在最上面加入,如果空间足够的话,然后优先放在最左面.统计每条公告最终的位置,即它所在的行数. 这里是线段树来存储 当前区间(i,j)的所有位置,剩余的最大空间. 初始化即为w,公告榜的宽. Problem Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h is…
Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15625    Accepted Submission(s): 6580 Problem Description At the entrance to the university, there is a huge rectangular billboard of…