题目:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2361 来源:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=26760#problem/B Beloved Sons Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Once upon a time there liv…

Fire Net Time Limit: 2 Seconds      Memory Limit: 65536 KB Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small c…

题目链接:Taxi Taxi Time Limit: 1 Second      Memory Limit: 32768 KB As we all know, it often rains suddenly in Hangzhou during summer time.I suffered a heavy rain when I was walking on the street yesterday, so I decided to take a taxi back school. I foun…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4836 因为要使对角线所有元素都是U,所以需要保证每行都有一个不同的列上有U,设(i,j)的位置是U, 以U为边,连接点i和点j+n,也即连接行点和列点,最大匹配为n则必定有解,否则必定无解 #include <cstdio> #include <iostream> #include <cstring> #include <cctype>…

题意: 给你一副图, 有草地(*),空地(o)和墙(#),空地上可以放机器人, 机器人向上下左右4个方向开枪(枪不能穿墙),问你在所有机器人都不相互攻击的情况下能放的最多的机器人数. 思路:这是一类经典题的衍化,如果没有墙,我们会将行和列看成两列点阵,然后就可以用二分匹配解. 现在有墙怎么办呢, 把某一行或列(有墙的拆分成多个区域,可以看成多个行或列), 拆好以后更没有墙的做法一样了. #include <cstdio> #include <cstring> #include &l…

The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24081   Accepted: 10695 Description Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering p…

刚回到家 开了二分匹配专题 手握xyl模板 奋力写写写 终于写完了一群模板题 A hdu1045 对这个图进行 行列的重写 给每个位置赋予新的行列 使不能相互打到的位置 拥有不同的行与列 然后左行右列 边是新的坐标 求最大匹配 #include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> #include<map> #include<string>…

1189: [HNOI2007]紧急疏散evacuate Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 1155  Solved: 420[Submit][Status][Discuss] Description 发生了火警,所有人员需要紧急疏散!假设每个房间是一个N M的矩形区域.每个格子如果是'.',那么表示这是一块空地:如果是'X',那么表示这是一面墙,如果是'D',那么表示这是一扇门,人们可以从这儿撤出房间.已知门一定在房间的边界上,并且…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5943 题意:给你两个数n, s 然后让你判断是否存在(s+1, s+2, s+3, ... , s+n )的任意排列方式使得每个数都满足当前数num,与num所在位置 pos  形成num%pos=0: 例如 n = 4 , s = 11 num = {13, 14, 15, 12} pos =  {1,    2,   3,    4} 每个num与之对应的pos都是num%pos = 0;的关系…

Taxi Cab Scheme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5710   Accepted: 2393 Description Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination of the cabs in order to pick up…