HDU--3466 Proud Merchants (01背包)】的更多相关文章

Proud Merchants Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1978    Accepted Submission(s): 792 Problem Description Recently, iSea went to an ancient country. For such a long time, it was…
题目链接: 传送门 Proud Merchants Time Limit: 1000MS     Memory Limit: 65536K Description Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are…
题目链接: 传送门 Proud Merchants Time Limit: 1000MS     Memory Limit: 65536K Description Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are…
题意:物品有三个属性,价格p,解锁钱数下线q(手中余额>=q才有机会购买该商品),价值v.钱数为m,问购买到物品价值和最大. 思路:首先是个01背包问题,但购买物品受限所以应先排序.考虑相邻两个物品1和2,假设先购买1再购买2,若…
HDU 3466 Proud Merchants 带有限制的01背包问题 题意 最近,伊萨去了一个古老的国家.在这么长的时间里,它是世界上最富有.最强大的王国.因此,即使他们的国家不再那么富有,这个国家的人民仍然非常自豪. 商人是最典型的,他们每一个只卖了一个项目,价格是PI,但他们拒绝与你交易如果你的钱低于QI,ISEA评估每一个项目的价值VI. 如果他有M单位的钱,伊萨能得到的最大价值是多少? 解题思路 p代表需要的价钱,q代表买这个物品的门槛. 思路就是先尝试买qi-pi大的商品,即按照q…
Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn't been so wealthy any more. The merchan…
Proud Merchants Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 4039    Accepted Submission(s): 1677 Problem Description Recently, iSea went to an ancient country. For such a long time, it was…
这道题目看出背包非常easy.主要是处理背包的时候须要依照q-p排序然后进行背包. 这样保证了尽量多的利用空间. Proud Merchants Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 2674    Accepted Submission(s): 1109 Problem Description Recently, iSea…
题意: 要买一些东西,每件东西有价格和价值,但是买得到的前提是身上的钱要比该东西价格多出一定的量,否则不卖.给出身上的钱和所有东西的3个属性,求最大总价值. 思路: 1)WA思路:与01背包差不多,dp过程中记录每个容量所能获得的最大价值以及剩余的容量.实现是,开个二维dp数组,从左往右扫,考虑背包容量为j的可获得价值量,根据该剩余容量得知要更新后面哪个背包容量[j+?] ,如果剩余容量大于q[i]那么可以直接装进去,更新价值以及剩余容量,否则,考虑更新的是更大的背包容量.这个思路有缺陷,一直找…
Proud Merchants Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 12   Accepted Submission(s) : 5 Problem Description Recently, iSea went to an ancient country. For such a long time, it was the mo…