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).
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.
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] 和 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：
