[HDU 2126] buy the souvenirs (Dynamic Planning)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2126

Question: give you n items, m yuan, and ask you how many items you can buy and how many solutions are available.

 

At first, I thought that it was a greedy algorithm to solve the problem of giving m yuan money because I could purchase enough items because I spent less money.

The maximum number of purchased items is C.

Then, the design status DP [I] [J] indicates how many solutions I had to buy C items when I spent J yuan in my previous items.

Later I found that the state dimension was not enough. I had to think about it again.

So I thought:

Design status DP [I] [J] [k] indicates that I bought J items from my previous items, and the amount of money spent cannot exceed K yuan.

State transition equation: DP [I] [J] [k] = DP [I-1] [J-1] [k-A [I] + dp [I-1] [J] [k]

That is, J items were bought from the previous I items, the amount of money spent does not exceed K yuan of the number of items in the former I-1 to buy J spend less than K Yuan and the former I-1 bought a J-1, the sum of the number of K-A [I] solutions.

Code:

 1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 #include <cmath> 5 #include <map> 6 #include <iterator> 7 #include <vector> 8 using namespace std; 9 typedef long long LL;10 11 int T,n,m;12 const int MAX_N = 33;13 int a[MAX_N];14 int dp[MAX_N][MAX_N][555];15 16 int main(){17     scanf("%d",&T);18     while(T--){19         scanf("%d%d",&n,&m);20         for(int i=1;i<=n;i++){21             scanf("%d",&a[i]);22         }23         memset(dp,0,sizeof(dp));24 //        dp[0][0][0] = 1;25         for(int i=0;i<=m;i++) dp[0][0][i] = 1;26         for(int i=1;i<=n;i++){27             for(int j=0;j<=i;j++){28                 for(int k=m;k>=0;k--){29                     dp[i][j][k] += dp[i-1][j][k];30                     if( k-a[i]>=0 ) dp[i][j][k] += dp[i-1][j-1][k-a[i]];31                 }32             }33         }34         int maxn = 0;35         for(int i=0;i<=n;i++){36             if( dp[n][i][m] ) maxn = i;37         }38         if( maxn ) printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n",dp[n][maxn][m],maxn);39         else puts("Sorry, you can‘t buy anything.");40     }41     return 0;42 }

 

[HDU 2126] buy the souvenirs (Dynamic Planning)