HDU 3578 Greedy Tino(ダブルタワーDP)

Greedy Tino
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1194    Accepted Submission(s): 393
Problem Description
  Tino wrote a long long story. BUT! in Chinese...
  So I have to tell you the problem directly and discard his long long story. That is tino want to carry some oranges with "Carrying pole", and he must make two side of the Carrying pole are the same weight. Each orange have its' weight. So greedy tino want to know the maximum weight he can carry.
 
Input
The first line of input contains a number t, which means there are t cases of the test data.
  for each test case, the first line contain a number n, indicate the number of oranges.
  the second line contains n numbers, Wi, indicate the weight of each orange
  n is between 1 and 100, inclusive. Wi is between 0 and 2000, inclusive. the sum of Wi is equal or less than 2000.
 
Output
For each test case, output the maximum weight in one side of Carrying pole. If you can't carry any orange, output -1. Output format is shown in Sample Output.
 
Sample Input

   
   
   
   

    
    
    
    1
5
1 2 3 4 5
   
   
   
   

 
Sample Output

   
   
   
   

    
    
    
    Case 1: 7
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
      DP
   
   
   
   
   
   
   
   

    
    
    
              0，       
   
   
   
   
   
   
   
   

    
    
    
    dp[i][j]    i   ，    
   
   
   
   
   
   
   
   

    
    
    
    

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
int n;
int a[105];
int dp[105][4005];
int main()
{
	int t;
	scanf("%d",&t);
	int cas=0;
	while(t--)
	{
		scanf("%d",&n);
		int num=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			if(a[i]==0)
				i--,n--,num++;
			
		}
		memset(dp,-1,sizeof(dp));
		dp[0][0+2000]=0;
		for(int i=1;i<=n;i++)
		{
			memcpy(dp[i],dp[i-1],sizeof(dp[i]));
			for(int j=-2000;j<=2000;j++)
			{
				
                if(dp[i-1][j+2000]==-1) continue;

				if(j<0)
				{
                    dp[i][j-a[i]+2000]=max(dp[i][j-a[i]+2000],dp[i-1][j+2000]+a[i]);
                    dp[i][j+a[i]+2000]=max(dp[i][j+a[i]+2000],dp[i-1][j+2000]+max(0,j+a[i]));
				}
				else
				{
					
					 dp[i][j+a[i]+2000]=max( dp[i][j+a[i]+2000],dp[i-1][j+2000]+a[i]);
                     dp[i][j-a[i]+2000]=max(dp[i][j-a[i]+2000],dp[i-1][j+2000]+max(0,a[i]-j));
				}
			}
		}
		
		if(dp[n][2000])
			printf("Case %d: %d
",++cas,dp[n][2000]);
		else
		{
			if(num>=1)
				printf("Case %d: %d
",++cas,0);
			else
			   printf("Case %d: %d
",++cas,-1);
		}
	}
	return 0;
}