HDU1059+多重背包

多重背包

题意：价值为1，2，3，4，5，6. 分别有n[1],n[2],n[3],n[4],n[5],n[6]个。

求能否找到满足价值刚好是所有的一半的方案。

View Code

 1 /*

 2 多重背包

 3 */

 4 #include<stdio.h>

 5 #include<stdlib.h>

 6 #include<string.h>

 7 #include<iostream>

 8 #include<algorithm>

 9 #include<queue>

10 #include<map>

11 #include<math.h>

12 using namespace std;

13 const int maxn = 10;

14 int n[ maxn ];

15 int sumv,V;

16 int dp[ 60011 ];

17 void bag01( int cost,int weight ){

18     for( int i=V;i>=cost;i-- ){

19         dp[ i ]=max(dp[ i ],dp[ i-cost ]+weight);

20     }

21     return ;

22 }

23 void bagall( int cost ,int weight ){

24     for( int i=cost;i<=V;i++ ){

25         dp[ i ]=max( dp[ i ],dp[ i-cost ]+weight);

26     }

27     return ;

28 }

29 

30 void mul_bag( int cost,int weight,int num ){

31     if( cost*num>=V ){

32         bagall( cost,weight );

33         return ;

34     }

35     int k=1;

36     while( k<=num ){

37         bag01( k*cost,k*weight );

38         num-=k;

39         k*=2;

40     }

41     bag01( num*cost,num*weight );

42     return ;

43 }

44 int main()

45 {

46     int i,k=1;

47     while(cin>>n[0]>>n[1]>>n[2]>>n[3]>>n[4]>>n[5],n[0]+n[1]+n[2]+n[3]+n[4]+n[5])

48     {

49         memset(dp,0,sizeof(dp));

50         sumv=n[0]+n[1]*2+n[2]*3+n[3]*4+n[4]*5+n[5]*6;

51         if(sumv%2==1)

52         {

53             printf("Collection #%d:\nCan't be divided.\n\n",k++);

54             continue;

55         }

56         V=sumv/2;

57         for(i=0;i<6;i++)

58         {

59             if(n[i]) mul_bag(i+1,i+1,n[i]);

60         }

61         if(V==dp[V]) printf("Collection #%d:\nCan be divided.\n\n",k++);

62         else printf("Collection #%d:\nCan't be divided.\n\n",k++);

63     }

64     return 0;

65 }