Nyist OJ 311 Full backpack (Dynamic programming Classic)

Full Backpack time limit: theMs | Memory Limit:65535KB Difficulty:4

Descriptive narrative

Just say test instructions, the complete backpack defines n items and a backpack with a capacity of V. An unlimited number of items are available for each item. The volume of the article I is C, the value is W.

The solution of which items are loaded into the backpack allows the volume of these items to be no more than the backpack capacity, and the value of the largest sum. The main requirement is that the backpack just fills the backpack, to find out the maximum value of the sum is how much.

Assuming you can't just fill the backpack, output no

Input

The

first line: N indicates how many sets of test data (N<7) are available.
Next, the first line of the test data for each group has two integer m. V. M indicates the number of items, and V indicates the total capacity of the backpack. (0<m<=2000,0<v<=50000)
The next M-line has two integers per line c,w the weight and value of each item (0<c<100000,0<w<100000)

Output

each set of test data output results (assuming that you can just fill the backpack, output full backpack when the maximum value of the items in the backpack sum.

Assuming it can't just fill the backpack, output No)

Example input

21 52 22 52 25 1

Example output

NO1

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Userid=acm_%e8%b5%b5%e9%93%ad%e6%b5%a9 "style=" text-decoration:none; Color:rgb (55,119,188) ">acm_ Zhao Minhao

Dynamic Programming Classic problems, there are several ideas, the best idea of the 01 knapsack problem of the second cycle of the order of a change, we get the best solution of the complete backpack;

This algorithm uses a one-dimensional array, first look at the pseudocode: (referring to the Backpack 9 said inside content)

For I=1..N

For V=0..V

F[v]=max{f[v],f[v-cost]+weight}

You will find that this pseudo-code is different from the pseudo-code of the 01 backpack only having a V loop order. Why is such a change feasible? First think about why P01 in accordance with V=V. 0 in reverse order to cycle.

This is due to the fact that the state in the first cycle of the f[i][v] is recursive by state f[i-1] [V-c[i]]. Other words. This is to ensure that each item is selected only once. Ensure that, in considering the strategy of "Select Item I", it is based on a sub-result (F[i-1][v-c[i]) that has not been selected for article I items. The full backpack now features an unlimited selection of items for each item. Therefore, when considering the strategy of "adding an item I", I am going to need a sub-result that may have been selected in item I (F[i][v-c[i]), so that it can and must be cycled in the order of V=0..V. This is the reason why this simple procedure is set up.

It is worth mentioning that the order of the two-layer for loop can be reversed in the pseudo-code above. This conclusion may lead to optimization of the algorithm's time constants.

This algorithm can also be derived from other ideas.

Like what. The state transfer equation for solving f[i][v-c[i] in the basic idea is explicitly written, substituting the original equation, and the equation can be found to be equivalent to the form:

f[i][v]=max{f[i-1][v],f[i][v-c[i]]+w[i]}

This equation is implemented with a one-dimensional array, and the pseudo-code above is obtained.

The following is the Code of implementation, the code for dynamic programming is very easy, and the most important is mastering the state transition equation:

#include <cstdio> #include <cstring> #define MAX (b) a>b?
A:bconst int maxn=50001;int Dp[maxn];int main () {    int n,m,v,i,j,c,w;    scanf ("%d", &n);    while (n--)    {      memset (dp,-10000,sizeof (DP)), it is also important to note that in the 01 backpack is initialized to 0, here to initialize a larger negative number      dp[0]=0;//here also pay attention to, Without this it would WA      scanf ("%d%d", &m,&v);      for (i=1;i<=m;i++)      {          scanf ("%d%d", &c,&w);          for (j=c;j<=v;j++)                Dp[j]=max (dp[j],dp[j-c]+w);//The state transfer equation is also consistent with 01 Knapsack      }      if (dp[v]<0) printf ("no\n") ;       else  printf ("%d\n", Dp[v]);    }    return 0;}

Nyist OJ 311 Full backpack (Dynamic programming Classic)