HDU 4292 Food (网络流,最大流) 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…

http://acm.hdu.edu.cn/showproblem.php?pid=4292 题意:和奶牛一题差不多,只不过每种食物可以有多种. 思路:因为食物多种,所以源点和汇点的容量要改下.还有Dinic又TLE了,用ISAP过. #include <cstdio> #include <algorithm> #include <iostream> #include <cstring> #include <string> #include &l…

http://acm.hdu.edu.cn/showproblem.php?pid=4292 给一些人想要的食物和饮料,和你拥有的数量,问最多多少人可以同时获得一份食物和一份饮料 写的时候一共用了2种拆点建图的方法... 1. 起点连接饮料和食物,容量为拥有的数量 每个人被拆成三个点,$a,b,c$  $a$被想要的食物连接,$b$被想要的饮料连接,$c$被$a,b$连接,容量均为1 然后$c$点连接汇点,容量为2,最后遍历所有静态链表节点,对于所有指向汇点的边,如果剩余容量为0,则答案++ 然…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=4292 Food Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4060    Accepted Submission(s): 1359 Problem Description You, a part-time dining service w…

很裸的一道最大流 格式懒得排了,注意把人拆成两份,一份连接食物,一份连接饮料 4 3 3 //4个人,3种食物,3种饮料 1 1 1 //食物每种分别为1 1 1 1 //饮料每种数目分别为1 YYN //第一个人对第1,2,3种食物的态度为接受,接受和拒绝 NYY YNY YNY YNY //第一个人对第1,2,3种饮料的态度为接受,拒绝和接受 YYN YYN NNY 3 #include<cstdio> #include<iostream> #include<algori…

Food Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3658    Accepted Submission(s): 1246 Problem Description You, a part-time dining service worker in your college’s dining hall, are now confus…

Food Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2136    Accepted Submission(s): 745 Problem Description You, a part-time dining service worker in your college’s dining hall, are now confuse…

题目大意:有N个人,然后有F种食品和D种饮料,每个人都有喜欢的饮料和食品,求出来这些食品最多能满足多少人的需求. 输入描述: 分析:以前是做过类似的题目的,不过输入的信息量比较大,还是使用邻接表的好些,邻接矩阵有可能会TLE,建图如下图流程 不过要注意人需要拆点的,因为一个人只食用一种食品就可以了,所以拆点让他们之间的流量为1.ps:很伤心的这题竟然做了6个小时,算是比较裸的最大流题目....最后终于检查出来错误的位置,dinic里面把while循环写成了if,貌似这是第二次犯这么二缺的事儿了.…

题目链接 给你f种食物, 以及每种食物的个数, d种饮料, 以及个数, n个人, 以及每个人可以接受的食物种类和饮料种类. 每个人必须得到一种食物和一种饮料. 问最后得到满足的人的个数. 因为一个人只能得到一种食物, 所以把人拆成两个点, 之间连一条权值为1的边. 建一个源点s, 汇点t, 每种食物向源点连边, 权值为食物的个数, 饮料向汇点连边, 权值为个数. 如果一个人可以接受某种饮料, u'就向饮料连一条权值为1的边: 如果一个人可以接受某种食物, 食物就向u连权值为1的边. 每次数组都要…

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 s…