HDU1171(01背包均分问题)】的更多相关文章

Big Event in HDU Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been spli…
Dividing coins Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eag…
Washing Clothes Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 9384   Accepted: 2997 Description Dearboy was so busy recently that now he has piles of clothes to wash. Luckily, he has a beautiful and hard-working girlfriend to help him…
#1038 : 01背包 时间限制:20000ms 单点时限:1000ms 内存限制:256MB 描述 且说上一周的故事里,小Hi和小Ho费劲心思终于拿到了茫茫多的奖券!而现在,终于到了小Ho领取奖励的时刻了! 小Ho现在手上有M张奖券,而奖品区有N件奖品,分别标号为1到N,其中第i件奖品需要need(i)张奖券进行兑换,同时也只能兑换一次,为了使得辛苦得到的奖券不白白浪费,小Ho给每件奖品都评了分,其中第i件奖品的评分值为value(i),表示他对这件奖品的喜好值.现在他想知道,凭借他手上的这…
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1171 题意:把商品分成两半,如不能均分,尽可能的让两个数相接近.输出结果:两个数字a,b且a>=b. 思路:01背包. 先把商品的总价值计算出来,sum/2做为背包的容量. 然后讲同种商品的多件,存储为不同商品 同样价值的形式,也就是我们用一个一维数组来存储,不用一个二维或是两个一维数组来存. 感想:好久没有做背包的题目了,今天来做,忘了好多思路,这提醒着我,学习不能一直都在学新东西,也要及时的复习.…
题目链接: id=3211">poj3211  hdu1171 这个题目比1711难处理的是字符串怎样处理,所以我们要想办法,自然而然就要想到用结构体存储.所以最后将全部的衣服分组,然后将每组时间减半,看最多能装多少.最后求最大值.那么就非常愉快的转化成了一个01背包问题了... . hdu1711是说两个得到的价值要尽可能的相等.所以还是把全部的价值分为两半.最后01背包,那么这个问题就得到了解决.. 题目: Washing Clothes Time Limit: 1000MS   Me…
题意:过山车有n个区域,一个人有两个值F,D,在每个区域有两种选择: 1.睁眼: F += f[i], D += d[i] 2.闭眼: F = F ,     D -= K 问在D小于等于一定限度的时候最大的F. 解法: 用DP来做,如果定义dp[i][j]为前 i 个,D值为j的情况下最大的F的话,由于D值可能会增加到很大,所以是存不下的,又因为F每次最多增加20,那么1000次最多增加20000,所以开dp[1000][20000],dp[i][j]表示前 i 个,F值为j的情况下最小的D.…
Team Them Up! Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7608   Accepted: 2041   Special Judge Description Your task is to divide a number of persons into two teams, in such a way, that: everyone belongs to one of the teams; every t…
传送门 Description Hasan and Bahosain want to buy a new video game, they want to share the expenses. Hasan has a set of N coins and Bahosain has a set of M coins. The video game costs W JDs. Find the number of ways in which they can pay exactly W JDs su…
题目链接: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1085 题意: 中文题诶~ 思路: 01背包模板题. 用dp[i][j]表示到第i个物品花去j空间能存储的最大价值, 那么很显然有 ; i<=n; i++){ ; j<=m; j++){ //注意这里的j是从0开始而非a[i] if(j>=a[i]){ dp[i][j]=max(dp[i-][j-a[i]]+b[i], dp[i-][j]); }els…