HDU 4292 Food

  • 2018-08-21
  • 10
  • 0

Description:

  You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.
  The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
  You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
  Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.

Input:

  There are several test cases.
  For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.
  The second line contains F integers, the ith number of which denotes amount of representative food.
  The third line contains D integers, the ith number of which denotes amount of representative drink.
  Following is N line, each consisting of a string of length F. e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.
  Following is N line, each consisting of a string of length D. e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.
  Please process until EOF (End Of File).

Output:

  For each test case, please print a single line with one integer, the maximum number of people to be satisfied.

Sample Input:

4 3 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY

Sample Output:

3

题目链接

将源点与食物建立流量为食物数量的边,将食物与人建立流量为1的边,将人与饮料建立流量为1的边,将饮料与汇点建立流量为饮料数量的边,跑最大流即可。

AC代码:

评论

还没有任何评论,你来说两句吧