POJ 2184（01背包）（负体积）

【POJ 2184（01背包）（负体积）】的更多相关文章

poj 2184 01背包变形【背包dp】

POJ 2184 Cow Exhibition Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14657 Accepted: 5950 Description "Fat and docile, big and dumb, they look so stupid, they aren't much fun..." - Cows with Guns by Dana Lyons The cows want to…

POJ 2184 01背包+负数处理

Cow Exhibition Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10200 Accepted: 3977 Description "Fat and docile, big and dumb, they look so stupid, they aren't much fun..." - Cows with Guns by Dana Lyons The cows want to prove to…

hdu 2184 01背包变形

转自:http://blog.csdn.net/liuqiyao_01/article/details/8753686 题意:这是又是一道01背包的变体,题目要求选出一些牛,使smartness和funness值的和最大,而这些牛有些smartness或funness的值是负的,还要求最终的smartness之和以及funness之和不能为负. 这道题的关键有两点:一是将smartness看作花费.将funness看作价值,从而转化为01背包:二是对负值的处理,引入一个shift来表示“0”,…

poj 1837 01背包

Balance Time Limit: 1000 MS Memory Limit: 30000 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main [Submit] [Status] [Discuss] Description Gigel has a strange "balance" and he wants to poise it. Actually, the device is different fr…

POJ 3628 01背包 OR 状压

思路: 1.01背包先找到所有奶牛身高和与B的差. 然后做一次01背包即可 01背包的容积和价格就是奶牛们身高. 最后差值一减输出结果就大功告成啦! 2. 搜索这思路很明了吧... 搜索的确可以过- 3. 模拟! 0到1< < n 来一遍.(状压呗) 01背包的: // by SiriusRen #include <cstdio> #include <algorithm> using namespace std; int N,B,sum=0,h[25],f[1000…

Proud Merchants（POJ 3466 01背包+排序）

Proud Merchants Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 4257 Accepted Submission(s): 1757 Problem Description Recently, iSea went to an ancient country. For such a long time, it was…

POJ 3624 01背包

初学DP,用贪心的思想想解题,可是想了一个多小时还是想不出. //在max中的两个参数f[k], 和f[k-weight[i]]+value[i]都是表示在背包容量为k时的最大价值 //f[k]是这个意思,就不用说了. //而f[k-weight[i]]+value[i]也表示背包容量为k时的最大价值是为什么呢? //首先,f[k-weight[i]]表示的是背包容量为k-weight[i]的容量,也就是说f[k-weight[i]] //表示的是容量还差weiht[i]才到k的价值,+walu…