HDU5616 Jam's balance 动态规划

Jam's balance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 774    Accepted Submission(s): 374

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 
    
    
    
    
     
   
   
   
   

 

题目大意：告诉你N个砝码，问能否测量出质量为M的物品。

分析：动态规划，对于前X个砝码，若能组合出质量为K的物品，则需要前X-1个砝码能组合出K-W[X]或K+W[X]或K的重量。

最后直接访问DP数组即可。

CODE：

//************************************************************************//
//*Author : Handsome How                                                 *//
//************************************************************************//
//#pragma comment(linker, "/STA    CK:1024000000,1024000000")
#pragma warning(disable:4996) 
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>                                                
#include <cassert>
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define fur(i,a,b) for(int i=(a);i<=(b);i++)
#define furr(i,a,b) for(int i=(a);i>=(b);i--)
#define cl(a) memset((a),0,sizeof(a))
using namespace std;
typedef long long LL;
//----------------------------------------------------
bool dp[25][2500];
int    a[30];
int main()
{
    //freopen("E:\\data.in", "r", stdin);
    ios :: sync_with_stdio(false);
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int n;
        scanf("%d", &n);
        fur(i, 1, n)scanf("%d", &a[i]);
        cl(dp);
        dp[0][0] = true;
        fur(i, 1, n)fur(j, 0, 2000) {
            if (j >= a[i])
                if (dp[i - 1][j - a[i]])
                    dp[i][j] = 1;
            if (j + a[i] <= 2000)
                if (dp[i - 1][j + a[i]])
                    dp[i][j] = 1;
            if (dp[i - 1][j])
                dp[i][j] = 1;
        }
        int q;
        scanf("%d", &q);
        while (q--)
        {
            int t;
            scanf("%d", &t);
            if (t > 2000) {
                printf("NO\n");
                continue;
            }
            if (dp[n][t]) {
                printf("YES\n");
                continue;
            }
            printf("NO\n");
        }
    }
    return 0;
}