HDU 3578 Greedy Tino（双塔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


双塔DP
注意如果有一个橘子是0，那么就符合条件
dp[i][j] 表示第i个橘子，两边差值

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