POJ 1698 Alice's Chance 题目链接 题意:拍n部电影.每部电影要在前w星期完毕,而且一周仅仅有一些天是能够拍的,每部电影有个须要的总时间,问能否拍完电影 思路:源点向每部电影连边,容量为d,然后每部电影相应能拍的那天连边,因为每天容量限制是1.所以进行拆点,然后连向汇点就可以 代码: #include <cstdio> #include <cstring> #include <queue> #include <algorithm> us…

将星期拆点,符合条件的连边,最后统计汇点流量是否满即可了,注意结点编号. #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<set> #include<map> #include<queue> #include<vector> #include<s…

id=1698" style="background-color:rgb(51,255,51)">主题链接 意甲冠军: 爱丽丝要拍电影.有n部电影,规定爱丽丝第i部电影在每一个礼拜仅仅有固定的几天能够拍电影,且仅仅能在前wi个周拍,而且这部电影要拍di天才干结束.问爱丽丝能不能拍全然部的电影 第一行代表有多少组数据,对于每组数据第一行代表有n部电影,接下来2到n+1行,每行代表一个电影,每行9个数,前面7个数,1代表拍.0代表不拍,第8个数代表要拍几天,第9个数代表有几…

题目链接: HDU:http://acm.hdu.edu.cn/showproblem.php?pid=4791 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5072 Problem Description Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her pr…

点击打开链接 Alice's Print Service Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1084    Accepted Submission(s): 240 Problem Description Alice is providing print service, while the pricing doesn't…

题目大意: 一个月饼店每一个小时做出月饼的花费不一样. 储存起来要钱.最多存多久.问你把全部订单做完的最少花费. 思路分析: ans = segma( num[]*(cost[] + (i-j)*s) ) 整理一下会发现式子就是 cost[]-j*s + i*s 对于每个订单,我们把i拿出来分析 所以也就用cost - j*s 建树. 然后在储存期间找到最小的花费即可了. #include <cstdio> #include <iostream> #include <algo…

Eliminate the Conflict Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1732    Accepted Submission(s): 751 Problem Description Conflicts are everywhere in the world, from the young to the elder…

Eliminate the Conflict Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1315    Accepted Submission(s): 563 Problem Description Conflicts are everywhere in the world, from the young to the elder…

Eliminate the Conflict Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1114 Accepted Submission(s): 468 Problem Description Conflicts are everywhere in the world, from the young to the elderly, fr…

5. 图的转换操作 图的转换操作主要有以下的方法: (1) Graph.mapVertices():对图的顶点进行转换,返回一张新图: (2) Graph.mapEdges():对图的边进行转换,返回一张新图. 代码: // 转换操作 println("*************************************************************") println("转换操作") println("**************…