Problem Description Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on a long thin stick). Have you, however, considered about the hardship of a kebab roaster while enjoying the delicious food? Well, here's a chance f…

Problem Description Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory…

HDU 2883 kebab 题目链接 题意:有一个烧烤机,每次最多能烤 m 块肉.如今有 n 个人来买烤肉,每一个人到达时间为 si.离开时间为 ei,点的烤肉数量为 ci,每一个烤肉所需烘烤时间为 di.注意一个烤肉能够切成几份来烤 思路:把区间每一个点存起来排序后.得到最多2 * n - 1个区间,这些就表示几个互相不干扰的时间,每一个时间内仅仅可能有一个任务器做.这样建模就简单了.源点连向汇点,容量为任务须要总时间,区间连向汇点,容量为区间长度.然后每一个任务假设包括了某个区间,之间就连…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2883 Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on a long thin stick). Have you, however, considered about the hardship of a kebab roaster while enjoying the delicio…

Problem Description Gabiluso is one of the greatest spies in his country. Now he's trying to complete an "impossible" mission ----- to make it slow for the army of City Colugu to reach the airport. City Colugu has n bus stations and m roads. Eac…

kebab Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 288364-bit integer IO format: %I64d      Java class name: Main Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on a long…

kebab Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1243    Accepted Submission(s): 516 Problem Description Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on…

Description Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others. Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although…

//////////在这幅图中我们首先要增广1->2->4->6,这时可以获得一个容量为2的流,但是如果不建立4->2反向弧的话,则无法进一步增广,最终答案为2,显然是不对的,然而如果建立了反向弧4->2,则第二次能进行1->3->4->2->5->6的增广,最大流为3. #include<stdio.h> #include<algorithm> #include<string.h> #include<q…

  HDU 1565 方格取数(1) 给你一个n*n的格子的棋盘,每个格子里面有一个非负数.从中取出若干个数,使得任意的两个数所在的格子没有公共边,就是说所取的数所在的2个格子不能相邻,并且取出的数的和最大. Input 包括多个测试实例,每个测试实例包括一个整数n 和n*n个非负数(n<=20) Output 对于每个测试实例,输出可能取得的最大的和 Sample Input 3 75 15 21 75 15 28 34 70 5 Sample Output 188直接用这个程序拿双倍经验吧~…