Tug of War（严格限制数量的二维费用背包）

Tug of War

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8500		Accepted: 2311

Description

A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.

Input

The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.

Output

Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.

Sample Input

3
100
90
200

Sample Output

190 200

Source

Waterloo local 2000.09.30

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int dp[45010][105],w[105];
int main()
{
    int i,j,k,n,sum,sum1,sum2,num;
    while(cin>>n)
    {
        sum=0;
        for(i=1;i<=n;i++)
        {
            cin>>w[i];
            sum+=w[i];
        }
        sum1=(sum+1)/2;
        num=(n+1)/2;
        memset(dp,0,sizeof(dp));
        //cout<<"ok"<<endl;
        for(i=1;i<=n;i++)
        {
            for(j=sum1;j>=w[i];j--)
            {
                for(k=num;k>=1;k--)
                {
                    dp[j][k]=max(dp[j][k],dp[j-w[i]][k-1]+w[i]);
                }
            }
        }
        if(dp[sum1][num]>sum-dp[sum1][num])
        {
            cout<<sum-dp[sum1][num]<<" "<<dp[sum1][num]<<endl;
        }
        else
        {
            cout<<dp[sum1][num]<<" "<<sum-dp[sum1][num]<<endl;
        }
    }
}

不知道怎么回事?用二维费用背包竟然过了，应该过不了！

5
50 2 25 21 2
正确答案应该是48 52

正确的应该是必须在n/2人之内找到重量不大于总重量一半的重量！

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int dp[45010][105],w[105];
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
    int i,j,k,n,sum,sum1,sum2,num,ans;
    while(cin>>n)
    {
        sum=0;
        for(i=1;i<=n;i++)
        {
            cin>>w[i];
            sum+=w[i];
        }
        sum1=(sum+1)/2;
        num=(n+1)/2;
        memset(dp,-1,sizeof(dp));
		dp[0][0]=0;
        //cout<<"ok"<<endl;
        for(i=1;i<=n;i++)
        {
            for(j=sum1;j>=w[i];j--)
            {
                for(k=num;k>=1;k--)
                {
					if(dp[j-w[i]][k-1]!=-1)
						dp[j][k]=max(dp[j][k],dp[j-w[i]][k-1]+w[i]);
                }
            }
        }
		ans=0;
		for(i=0;i<=sum1;i++)
		{
			if(dp[i][num]>ans)
			{
				ans=dp[i][num];
			}
			if(dp[i][num-1]>ans)
			{
				ans=dp[i][num-1];
			}
		}
		if(ans<sum-ans)
        {
            cout<<ans<<" "<<sum-ans<<endl;
        }
        else
        {
            cout<<sum-ans<<" "<<ans<<endl;
        }
    }
}