Retrospective method 01 Knapsack Problem __ Backtracking method

problem

01 knapsack problem before using dynamic programming method to achieve: Dynamic programming 01 knapsack problem (easy to understand) 01 knapsack problem general Use the subset tree frame of backtracking method to realize

#include <iostream> using namespace std; int n,c,bestp;//items number, backpack capacity, maximum value int p[10000],w[10000],x[10000],bestx[10000];//items value, weight of items, selection of goods//CP current total value,  
            CW currently put in total weight void backtrack (int i,int cp,int CW) {if (i>n) {if (CP&GT;BESTP) {  
            BESTP=CP;  
        for (i=1;i<=n;i++) bestx[i]=x[i];  
            If {//every layer of items can be selected to put or not put, 0 means not put, 1 means put in for (int j=0;j<=1;j++) {  
            X[i]=j;  
                if (cw+x[i]*w[i]<=c) {cw+=w[i]*x[i];  
                Cp+=p[i]*x[i];  
                Backtrack (I+1,CP,CW);  
                Cw-=w[i]*x[i];  
            Cp-=p[i]*x[i];   
    int main () {bestp=0;  
    cin>>n>>c;  
    for (int i=1;i<=n;i++) cin>>w[i];  
    for (int i=1;i<=n;i++) cin>>p[i];  
    Backtrack (1,0,0);  
    cout<<bestp<<endl;for (int i=1;i<=n;i++) cout << bestx[i] << ""; }

Input:
3 10
5 5 3
40 30 10

Output:
70
1 1 0
Process Analysis:
If you use an array to represent each recursive, the choice of each item, then: (0 for the election, 1 for the election)
(0,0,0) All BESTP = 0
(0,0,1) Select 3rd Item BESTP = 10
(0,1,0) .....
(0,1,1) .....
(1,0,0) .....
(1,0,1) .....
(1,1,0) Select 1th and 2nd, the weight is 10,bestp=70

(1,1,1) All options, weight of 13, cut off of course can also be used to arrange the tree to achieve

#include <iostream> using namespace std; int n, c, Bestp=0, cw=0, cp=0;  Number of items, backpack capacity, maximum value, current selected weight, currently selected value int w[10], v[10], x[10], bestx[10];//value of items, weight of items, selection of items int vis[10];//record whether int tol =
    0;//records recursive number of times void back (int cur) {tol++;
            if (cur > N) {if (cp > BESTP) {BESTP = CP;
            for (int i=1; i<=n; i++) {bestx[i] = X[i];
                }}else{for (int i=1; i<=n; i++) {if (!vis[i] && cw+w[i] <= c) {
                Vis[i] = 1;
                CW + = W[i];
                X[cur] = W[i];
                CP = V[i];

                Back (cur+1);
                Reset CW-= W[i];
                CP-= V[i]; 
                X[cur] = 0;
            Vis[i] = 0;
            } if (Cw+w[i] > C) {back (cur+1);
    int main () {CIN >> n >> C; for (int i = 1; I <= n; i++) {cin >>W[i];
    for (int i = 1; I <= n; i++) {cin >> v[i];
    } back (1);
    cout << BESTP << Endl;
    for (int i = 1; I <= n; i++) {cout << bestx[i] << "";
} cout << Endl << tol;  }

Input:
3 10
5 5 3
40 30 10

Output:
70
5 5 0