Guardian of Decency Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 5244   Accepted: 2192 Description Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is afraid that…

题目链接:http://poj.org/problem?id=2771 Guardian of Decency Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 5517   Accepted: 2322 Description Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on…

Problem DescriptionIn the second year of the university somebody started a study on the romantic relations between the students. The relation "romantically involved" is defined between one girl and one boy. For the study reasons it is necessary…

Problem DescriptionFrank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is afraid that some of them might become couples. While you can never exclude this possibility, he has made so…

题意: 有G个女孩,B个男孩.女孩彼此互相认识,男孩也彼此互相认识.有M对男孩和女孩是认识的.分别是(g1,b1),.....(gm,bm). 现在老师要在这G+B个小孩中挑出一些人,条件是这些人都互相认识.问最多可以挑出多少人. 思路: 女孩之间互相认识,男孩之间互相认识,所以我们可以将连边定义为:不认识.即:若两个节点之间有连边,则两个节点互不认识. 故题意即为选出最多的点使得这些点任意两点之间没有连边.即选最少的点覆盖所有边.(二分图最大独立集/二分图最小点覆盖) 代码: vector<i…

poj——2771    Guardian of Decency Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 5916   Accepted: 2458 Description Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he i…

题意:动物园有n只猫和m条狗,现在有p个小孩,他们有的喜欢猫,有的喜欢狗,其中喜欢猫的一定不喜欢狗,喜欢狗的一定不喜欢猫.现在管理员要从动物园中移除一些动物,如果一个小孩喜欢的动物留了下来而不喜欢的动物被移走,这个小孩会很高兴.现在问最多可以让多少个小孩高兴. 此题是求二分图最大独立集. 二分图比较明显,但是难在建图.这个题是找到最多的喜欢猫和喜欢狗而不互相冲突的小孩,这样我们将喜欢动物相互冲突的小孩之间连边,问题就变成了求二分图的最大独立集. 在二分图中,最大独立集=顶点数-最大匹配数. 求解…

Cat VS Dog Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 4039    Accepted Submission(s): 1458 Problem Description The zoo have N cats and M dogs, today there are P children visiting the zoo,…

Description 给定一个01矩阵,其中你可以在0的位置放置攻击装置.每一个攻击装置(x,y)都可以按照“日”字攻击其周围的 8个位置(x-1,y-2),(x-2,y-1),(x+1,y-2),(x+2,y-1),(x-1,y+2),(x-2,y+1), (x+1,y+2),(x+2,y+1) 求在装置互不攻击的情况下,最多可以放置多少个装置. Input 第一行一个整数N,表示矩阵大小为N*N.接下来N行每一行一个长度N的01串,表示矩阵. Output 一个整数,表示在装置互不攻击的情…

传送门 模型 二分图最大独立集,转化为二分图最大匹配,从而用最大流解决. 实现 首先把棋盘黑白染色,使相邻格子颜色不同. 把所有可用的黑色格子看做二分图X集合中顶点,可用的白色格子看做Y集合顶点. 建立附加源S汇T,从S向X集合中每个顶点连接一条容量为1的有向边,从Y集合中每个顶点向T连接一条容量为1的有向边. 从每个可用的黑色格子向骑士一步能攻击到的可用的白色格子连接一条容量为无穷大的有向边. 求出网络最大流,要求的结果就是可用格子的数量减去最大流量. 分析 用网络流的方法解决棋盘上的问题,一…