1171 Big Event in HDU【背包】

Big Event in HDU

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

Total Submission(s): 30040 Accepted Submission(s): 10543

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

给出物品的价值和每个物品的数量，求得在不超过一定容积的限定下，价值最大........

这个题相当于求容积只有一半的时候的背包问题....

跑的特别慢........

/*
未进行特殊处理... 当成多重背包
*/
#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
int dp[250005],val[105],num[105],n,sum;
void slove()
{
	memset(dp,0,sizeof(dp));
	for(int i=0;i<n;++i)//某一个
	{
		for(int j=1;j<=num[i];++j)//物品个数
		{
			for(int k=sum/2;k>=j*val[i];--k)//容积
			{
				dp[k]=max(dp[j],dp[k-val[i]]+val[i]);
			}
		}
	}
	printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);//最终
}
int main()
{
	//freopen("shuju.txt","r",stdin);
	while(~scanf("%d",&n),n>0)
	{
		sum=0;
		for(int i=0;i<n;++i)
		{
			scanf("%d%d",val+i,num+i);
			sum+=val[i]*num[i];
		}
		slove();
	}
	return 0;
}

特殊处理成01 背包.......

/*
01背包 
*/
#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
int dp[250005],val[10005],n,sum,cnt;
int main()
{
	while(~scanf("%d",&n),n>0)
	{
		sum=0,cnt=0;
		for(int i=0;i<n;++i)
		{
			int vi,num;
			scanf("%d%d",&vi,&num);
			sum+=vi*num;
			for(int j=0;j<num;++j)
			{
				val[cnt++]=vi;
			}
		}
		memset(dp,0,sizeof(dp));
		for(int i=0;i<cnt;++i)
		{
			for(int j=sum/2;j>=val[i];--j)
			{
				dp[j]=max(dp[j],dp[j-val[i]]+val[i]);
			}
		}
		printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
	}
	return 0;
}

其实并没有多少改进，见到别人0ms 过的，真心是不懂他们的方法了..............

慢慢学........