0-1 knapsack problem

Questions:

Given n kinds of goods and a backpack. The weight of item I is WI, its value is VI, the capacity of the backpack is C.

For an item that is only loaded or not installed, the problem is called a 0-1 knapsack problem.

Ask how to choose the items in your backpack so that the total value of the items loaded into the backpack is the greatest.

Analysis:

n Kinds of Goods: 1 2 3 ..... ... n

　　Use M (i, J) to indicate an optional item for I, I+1, ..., n while the backpack capacity is only the value of the item in the backpack when J is the best choice.

At this point, the M (I, J) recursion is defined as follows:

　　

That is to say, for the whole problem, if the capacity of the next I, see the next item I pack the value of the backpack, or not pack the value of large (I or not to affect other items).  

　　

Algorithm Ideas:

We calculate the nth item first, and then calculate the n-1 until the top one.

Code: 

　　

voidKnapack (intV[],intW[],intCintNint**m) {    intJmax = min (W[n]-1, c);  for(intj =0; J <= Jmax; J + +)//if the backpack capacity is less than n size{M[n][j]=0;//it will not fit N    }     for(intj = W[n]; J <= C; J + +)//if the backpack capacity is greater than the size of n{M[n][j]= V[n];//it's loaded with N.    }     for(inti = n-1; I>1; i--) {Jmax= Min (W[i]-1, c);  for(intj =0; J <= Jmax; J + +)//if the backpack capacity is less than the size of I{M[i][j]= M[i +1][J];//I cannot be installed, I have to continue to install the back        }         for(intj = W[i]; J <= C; J + +)//if the backpack capacity is larger than the size of I{M[i][j]= max (M[i +1][J], M[i +1][j-w[i]] + v[i]);//look at the end of the pack I will be better, or not installed better        }    }    //We need special consideration here .m[1][C] = m[2][c]; if(c >= w[1]) {m[1][C] = max (m[1][c], m[2][c-w[1]] + v[1]); }}voidTraceback (intm[][Ten],intW[],intCintNintx[]) {       for(intI=1; i<n; i++)      {          if(M[i][c] = = m[i+1][c]) {X[i]=0; }          Else{X[i]=1; C-=W[i]; }} X[n]= (M[n][c])?1:0; }  

0-1 knapsack problem