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): 24316    Accepted Submission(s): 8053 Problem Description The famous ACM (Advanced Computer Maker) Company has rented a floor of a b…

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…

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…

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…

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…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1050 AC code: #include<stdio.h> #include<string.h> int m[210]; int main(void) { int t,n,i,j,st,en,ma; scanf("%d",&t); while(t--) { memset(m,0,sizeof(m)); scanf("%d",&n)…

http://acm.hdu.edu.cn/showproblem.php?pid=1050 这个题我首先直接用的常规贪心,用的和那个尽可能看更多完整节目那种思路.但是.......一直WA....T_T.... 后来在网上搜了一下这个题,发现好多人都有问题,都没有求出来,基本上都用的对尾部排序求的方法. 其实这个题因为是两排房间,所以1和2公用一个走廊,其中一个在需要移动的时候宁外一个还是不能移动. 所以我后面改了思路,直接改成了用两次排序直接找里面重叠部分最多的.(尾部排序的时候也要处理走廊…