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…

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…

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…

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 for you to help the p…

题意: 有n个人去撸串,每个人都能决定自己的串上有几块肉,每一块肉都要花费一个单位时间才熟,烤炉一次能烤m块肉 给出每个人的起始时间.终止时间.要几串.每个串上有几块肉,问能否满足所有的人 (啥?题不是这么说的?...这样看就对了...) 解析: 和hdu3572一样 只不过这个时间的范围比较大 所以就不能以时间点建点了  所以要以时间段建点 把所有的时间放到数组里,这里我用的是vector,然后排序,去重,对于每个人 遍历所有的时间,如果有个时间段在当前人购买的时间里  就把人和这个时间段连边…

kebab Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1603    Accepted Submission(s): 677 Problem Description Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on…

建图           源点  ->     每个人  ->           每段时间      ->      汇点 时间要离散化一下 分成一些时间段 权                     inf    ti[i]*(time[i]-time[i-1])  m*(time[i]-time[i-1]) n人 m个机器 开始时间  总共要的数目 底线时间  每个所需时间 如果可以达到输出Y 否则N #include<stdio.h> #include<algo…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…