zoukankan      html  css  js  c++  java
  • HDU 1171 Big Event in HDU

    Big Event in HDU

    Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 8035 Accepted Submission(s): 2697


    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

    转换一下,half= 全部物体体积的一半,就可以直接套用多重背包, 输出: 全部物体的体积 - fine[half]fine[half]

     1 #include<stdio.h>
    2 #include<string.h>
    3 int fine[100000] , value[55] , num[55] ;
    4 int sum , n , half ;
    5 int main ()
    6 {
    7 while ( scanf ( "%d" , &n ) , n >= 0 )
    8 {
    9 sum = 0 ;
    10 for ( int i = 0 ; i < n ; i ++ )
    11 {
    12 scanf ( "%d%d" , &value[i] , &num[i] ) ;
    13 sum += ( value[i] * num[i] ) ;
    14 }
    15 half = sum / 2 ;
    16 for ( int i = 0 ; i <= half ; i ++ )
    17 fine[i] = 0 ;
    18 for ( int i = 0 ; i < n ; ++ i )
    19 for ( int j = 0 ; j < num[i] ; ++ j )
    20 for ( int k = half ; k >= value[i] ; -- k )
    21 if ( fine[k] < fine[k - value[i]] + value[i] )
    22 fine[k] = fine[k - value[i]] + value[i] ;
    23 printf ( "%d %d\n" ,sum - fine[half] , fine[half] ) ;
    24
    25 }
    26 return 0 ;
    27 }

      当然是多重背包则我们可以用二进制优化:

    http://www.cnblogs.com/jbelial/articles/2116565.html

  • 相关阅读:
    如何使用IntelliJ IDEA 14创建基于Maven3的Java Web Project
    一些python语法的合集
    thuwc2019滚粗记
    noip2018总结
    树链剖分
    NOIp2017游记(滚粗之旅)
    使用Redux管理你的React应用
    校招面试中积累的前端问题
    WebService 生成类的命令语句
    Request 请求页面的地址路径获取
  • 原文地址:https://www.cnblogs.com/jbelial/p/2116562.html
Copyright © 2011-2022 走看看