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…

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…

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…

In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5472    Accepted Submission(s): 1843 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project t…

题意:给你若干个集合,每个集合内的物品要么选任意一个,要么所有都选,求最后在背包能容纳的范围下最大的价值. 分析:对于每个并查集,从上到下滚动维护即可,其实就是一个01背包= =. 代码如下: #include <stdio.h> #include <algorithm> #include <string.h> #include <vector> using namespace std; + ; int w[N],b[N]; int n,m,W; int r…

Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 34532   Accepted: 15301 Description Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible fro…

http://poj.org/problem?id=2184   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 the public that they are both smart and fun. In order to do this,…

/* 题意:有 n 个站点(编号1...n),每一个站点都有一个能量值,为了不让这些能量值连接起来,要用 坦克占领这个站点!已知站点的 之间的距离,每个坦克从0点出发到某一个站点,1 unit distance costs 1 unit oil! 最后占领的所有的站点的能量值之和为总能量值的一半还要多,问最少耗油多少! */ /* 思路:不同的坦克会占领不同的站点,耗油最少那就是路程最少,所以我们先将从 0点到其他各点的 最短距离求出来!也就是d[i]的值!然后我们又知道每一个站点的所具有的能量…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5410 Problem Description Today is CRB's birthday. His mom decided to buy many presents for her lovely son.She went to the nearest shop with M Won(currency unit).At the shop, there are N kinds of pr…

题目链接 http://acm.hust.edu.cn/vjudge/contest/130883#problem/C Problem Description Zero Escape, is a visual novel adventure video game directed by Kotaro Uchikoshi (you may hear about ever17?) and developed by Chunsoft. Stilwell is enjoying the first ch…

Washing Clothes Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 9707   Accepted: 3114 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…

描述 XC的儿子小XC最喜欢玩的游戏用积木垒漂亮的城堡.城堡是用一些立方体的积木垒成的,城堡的每一层是一块积木.小XC是一个比他爸爸XC还聪明的孩子,他发现垒城堡的时候,如果下面的积木比上面的积木大,那么城堡便不容易倒.所以他在垒城堡的时候总是遵循这样的规则. 小XC想把自己垒的城堡送给幼儿园里漂亮的女孩子们,这样可以增加他的好感度.为了公平起见,他决定把送给每个女孩子一样高的城堡,这样可以避免女孩子们为了获得更漂亮的城堡而引起争执.可是他发现自己在垒城堡的时候并没有预先考虑到这一点.所以他现在…

2427: [HAOI2010]软件安装 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 960  Solved: 380[Submit][Status][Discuss] Description 现在我们的手头有N个软件,对于一个软件i,它要占用Wi的磁盘空间,它的价值为Vi.我们希望从中选择一些软件安装到一台磁盘容量为M计算机上,使得这些软件的价值尽可能大(即Vi的和最大).但是现在有个问题:软件之间存在依赖关系,即软件i只有在安装了软件j(包…

题意:求解01背包价值的第K优解. 分析: 基本思想是将每个状态都表示成有序队列,将状态转移方程中的max/min转化成有序队列的合并. 首先看01背包求最优解的状态转移方程:\[dp\left[ j \right] = \max \left\{ {dp\left[ j \right],dp\left[ {j - a\left[ i \right].w} \right] + a\left[ i \right].v} \right\}\] 如果要求第K优解,那么状态 dp[j] 就应该是一个大小为…

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …The bone collector had a big bag with a volume of V ,and along his tri…

    牛的展览会 题目大意:Bessie要选一些牛参加展览,这些牛有两个属性,funness和smartness,现在要你求出怎么选,可以使所有牛的smartness和funness的最大,并且这两个和都不能为负值 这一题很有意思,首先是这个问题是二维的,它包含两个属性,但是他有一个很重要的条件就是牛只能选一次,所以我们一开始就很容易想到用背包貌似可以求解,但是这一题没有办法直接用背包,因为没有直接给出价值和容量(他给了两个价值). 但是,我们稍微变换一下,这一题就可以做了,我们要把一个看成是…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=3448 Description 0/1 bag problem should sound familiar to everybody. Every earth man knows it well. Here is a mutant: given the capacity of a bag, that is to say, the number of goods the bag can ca…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=3496 //刚看题目以为是简单的二维01背包,but,,有WA点.. 思路:题中说,只能买M个光盘,不能多也不能少,所以就要求把背包装满. 恰好把背包装满,那么在初始化时,除了dp[0]=0,剩下的dp[1~M],均为负无穷(其实设置成-1,到时候在判断一下也是一样的,思想相同) 这样才可以保证最终得到的dp[M]是一种恰好装满背包状态的最优解. 代码: #include<iostream…

这题和NOIP的金明的预算方案(?)很像,只不过附件的数量增多了 如果对主件进行一次01背包,再套一层附件的01背包O(n4)肯定会爆.. 所以我们可以先预处理出,对于每个主件,花的时间为k的情况下,最大的经验值,用01背包做 然后再对每个主件进行01背包,这样就去掉了一层循环 #include<stdio.h> #include<string.h> #include<algorithm> #define maxn 102 using namespace std; ],…

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…

Robberies  HDU2955 因为题目涉及求浮点数的计算:则不能从正面使用01背包求解... 为了能够使用01背包!从唯一的整数(抢到的钱下手)... 之后就是概率的问题: 题目只是给出被抓的几率,如果同时抢两家银行的话,那么被抓的概率是: (1-一家不被抓的概率*另一家不被抓的概率) 才是同时抢两家被抓的概率! 最后和题目给出的概率比较取较大值... 那么赋初值的时候dp[0]=1. 注意:不要误以为精度只有两位. #include<iostream> #include<std…

http://acm.hdu.edu.cn/showproblem.php?pid=1864 New~ 欢迎“热爱编程”的高考少年——报考杭州电子科技大学计算机学院关于2015年杭电ACM暑期集训队的选拔 最大报销额 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18562    Accepted Submission(s): 5459…

http://acm.hdu.edu.cn/showproblem.php?pid=2955 如果认为:1-P是背包的容量,n是物品的个数,sum是所有物品的总价值,条件就是装入背包的物品的体积和不能超过背包的容量1-P. 在这个条件下,让装入背包的物品的总价值,也就是bag[i].[v]的和最大 bag.v是每一件物品的价值,bag.p是每件物品的体积 像上面这样想是行不通的.下面有解释 这道题麻烦的是概率这东西没法用个循环表示出来,根据我以往的经验,指望着把给出的测试数据乘上一百或者一万这种…

题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=203 思路:先求点0到每个点的最短距离,dijkstra算法,然后就是01背包了 我奇怪的是100*1000000的时间复杂度居然没有超时! 代码如下: #include "stdio.h" #include "string.h" #define N 105 #define INF 0x3fffffff bool mark[N]; int dist[N];…