Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 36096    Accepted Submission(s): 12533

Problem 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 split into Computer College and Software College in 2002.

The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is
N (0<N<1000) kinds of facilities (different value, different kinds).
 
Input
Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the
facilities) each. You can assume that all V are different.

A test case starting with a negative integer terminates input and this test case is not to be processed.
 
Output
For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.
 
Sample Input
2
10 1
20 1
3
10 1
20 2
30 1
-1
 
Sample Output
20 10
40 40
 
Author
lcy
 
题目大意:
相当于一堆东西要分到两个篮子A和B中,让B中的价值量尽可能地靠近A但不能超过A。

解题思路:
无限靠近但不能超过,就是以总价值量的一半作为划分的界限。然后就拆解为01背包的问题了,背包大小是总背包的一半,往B的篮子中装,求能够装载的最大价值量。题目中给定的是物品的价值和对应的数量,转化为01背包,要变成一个个的单个物体,不能像题目中依靠价值的不同划分为不同的类别。如果划分为单个物品,极限考虑就是50种物品,物品的价值从1到50,每种物品有100个,得到的总背包大小就是(50+1)*50/2*100=1275000。同时,一般的01背包会有体积的说法,这里的话可以把物品的价值看作物品的体积,自己体会一下,比较容易懂。还有就是输入-1结束,不能写成
if(n == -1) return 0;因为这个WA很多次。改成if(n < 0) return 0;就A了。郁闷(-__-)b

源代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<deque>
#include<map>
#include<set>
#include<algorithm>
#include<string>
#include<iomanip>
#include<cstdlib>
#include<cmath>
#include<sstream>
#include<ctime>
using namespace std; int va[1275005];//(50+1)*50/2*100=1275000最大价值作为背包的最大空间
int dp[1275005]; int main()
{
int n;
int i,j;
int v,num;//价值,数量
int count_num;//所有物品的数量
int count_pac;//总的背包大小
while(scanf("%d",&n) != EOF)
{
if(n < 0)
{
return 0;
}
memset(va,0,sizeof(va));
memset(dp,0,sizeof(dp));
j = 1;
count_num = 0;
count_pac = 0;
for(i = 1; i <= n; i++)
{
scanf("%d%d",&v,&num);
count_num += num;
count_pac += num * v;
while(num--)
{
va[j] = v;
j++;
}
}
//01背包
for(i = 1; i <= count_num; i++)
{
for(j = count_pac / 2; j >= va[i]; j--)
{
dp[j] = max(dp[j], dp[j - va[i]] + va[i]);
}
}
printf("%d %d\n",count_pac - dp[count_pac / 2], dp[count_pac / 2]);
}
return 0;
}

HDU1171--01背包的更多相关文章

  1. HDU1171(01背包均分问题)

    Big Event in HDU Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u De ...

  2. poj3211Washing Clothes(字符串处理+01背包) hdu1171Big Event in HDU(01背包)

    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 ...

  3. UVALive 4870 Roller Coaster --01背包

    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 ...

  4. POJ1112 Team Them Up![二分图染色 补图 01背包]

    Team Them Up! Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7608   Accepted: 2041   S ...

  5. Codeforces 2016 ACM Amman Collegiate Programming Contest A. Coins(动态规划/01背包变形)

    传送门 Description Hasan and Bahosain want to buy a new video game, they want to share the expenses. Ha ...

  6. 51nod1085(01背包)

    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[ ...

  7. *HDU3339 最短路+01背包

    In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  8. codeforces 742D Arpa's weak amphitheater and Mehrdad's valuable Hoses ——(01背包变形)

    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: #in ...

  9. POJ 3624 Charm Bracelet(01背包)

    Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 34532   Accepted: 15301 ...

  10. (01背包变形) Cow Exhibition (poj 2184)

    http://poj.org/problem?id=2184   Description "Fat and docile, big and dumb, they look so stupid ...

随机推荐

  1. LeetCode 598. Range Addition II (范围加法之二)

    Given an m * n matrix M initialized with all 0's and several update operations. Operations are repre ...

  2. thinkphp3.2开发网页实现第三方登录

    1.在要添加登录的html里添加登录按钮: <a href="{:U('Public/login/',array('type'=>'weixin'))}">< ...

  3. 【Salvation】——关卡功能&数据库基础实现

    写在前面:项目的关卡功能和数据库基础实现是小组其他成员实现的部分,这里作为学习总结.关卡功能块使用C#语言编写脚本,在Unity3D游戏引擎的环境中实现,数据库功能块使用PHP作为服务端获取MySQL ...

  4. (2017浙江省赛E)Seven Segment Display

    Seven Segment Display Time Limit: 2 Seconds      Memory Limit: 65536 KB A seven segment display, or ...

  5. 链表倒数第n个节点

    找到单链表倒数第n个节点,保证链表中节点的最少数量为n. 样例 给出链表 3->2->1->5->null和n = 2,返回倒数第二个节点的值1. /** * Definiti ...

  6. html5页面实现点击复制功能

    在实际工作中,有时候会遇到这样的需求,页面上有一个链接,不需要选中链接内容,只需要点击复制按钮,就可以把链接内容复制到剪切板.这时候可以使用clipboard插件来实现.以下是一个简单的demo. 首 ...

  7. spring boot hello and docker

    主要是想试下spring boot运行在docker里的感觉, 小试牛刀   :) 这是原文,参考一下:  https://spring.io/guides/gs/spring-boot-docker ...

  8. Python爬虫入门:综述

    大家好哈,最近博主在学习Python,学习期间也遇到一些问题,获得了一些经验,在此将自己的学习系统地整理下来,如果大家有兴趣学习爬虫的话,可以将这些文章作为参考,也欢迎大家一共分享学习经验. Pyth ...

  9. 【Kafka源码】KafkaController启动过程

    [TOC] 之前聊过了很多Kafka启动过程中的一些加载内容,也知道了broker可以分为很多的partition,每个partition内部也可以分为leader和follower,主从之间有数据的 ...

  10. oracle数据库管理系统常见的错误(一)

    oracle数据库管理系统常见的错误之一如下: Listener refused the connection with the following error:ORA-12519, TNS:no a ...