HDU 5616 Jam's balance（dp）

Jam's balance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 356    Accepted Submission(s): 163

Problem Description

Jim has a balance and N weights.  (1≤N≤20)
The balance can only tell whether things on different side are the same weight.
Weights can be put on left side or right side arbitrarily.
Please tell whether the balance can measure an object of weight M.

 

Input

The first line is a integer  T(1≤T≤5) , means T test cases.
For each test case :
The first line is  N , means the number of weights.
The second line are  N  number, i'th number  wi(1≤wi≤100)  means the i'th weight's weight is  wi .
The third line is a number  M .  M  is the weight of the object being measured.

 

Output

You should output the "YES"or"NO".

 

Sample Input

   
   
   
   

    
    
    
    1
2
1 4
3
2
4
5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    NO
YES
YES


    
    
    
    
 
     
 
      Hint 
     
 For the Case 1:Put the 4 weight alone For the Case 2:Put the 4 weight and 1 weight on both side 
    
    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    思路:dp【i】【j】表示选到第i个砝码，可以称重重量为j的物品的状态，只有0或1代表YES和NO，然后状态转移，每个砝码可以不选，和物品放一边，不和物品放一边三种选择，然后状态转移，预处理好dp数组后，每个查询都是O(1)了。
   
   
   
   
   
   
   
   

    
    
    
    
<pre name="code" class="cpp">#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
using namespace std;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f, N = 6e4;
typedef long long ll;
using namespace std;
int main()
{
    int w[22], dp[22][2020], sum;
    int t, n, m, x;
    cin>>t;
    while(t--)
    {
        sum = 0;
        scanf("%d", &n);
        for(int i =  1; i<=n; i++)
            scanf("%d", &w[i]), sum += w[i];
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for(int i = 1; i<=n; i++)
            for(int j = 0; j<=sum; j++)
            {
                dp[i][j] |= dp[i-1][j];
                dp[i][j] |= dp[i-1][abs(j-w[i])];
                if(j + w[i]<=sum)
                    dp[i][j] |= dp[i-1][j+w[i]];
            }
        scanf("%d", &m);
        while(m--)
        {
            scanf("%d", &x);
            if(dp[n][x] && x <= sum && x > 0)
                puts("YES");
            else puts("NO");
        }
    }
    return 0;
}