By zeyad_alaa, history, 11 months ago,

Hello so this is my first blog, I solved knapsack 1 but cant find any solution to solve V2 : https://atcoder.jp/contests/dp/tasks/dp_e

can any one help me ?

• +20

 » 11 months ago, # | ← Rev. 2 →   +2 first you have to find value per weight,then you sort them in descending order. then take one by one
•  » » 10 months ago, # ^ |   0 this didn't work for me. It gave correct answer for a few cases, but then it failed 3/4th of the cases.Could you explain why?
 » 11 months ago, # |   +10 This is a variant of the classic knapsack.Notice that value is small (at most $10^5$ for all $100$ items), so instead of "what is the most value we can carry with weight $W$", you find "what is the least weight you need to carry to get a value of $V$".
•  » » 4 months ago, # ^ |   0 Can it be done using top-down approach?
•  » » » 4 months ago, # ^ |   0 I don't see why not :)The parameters are (the item we are considering, the value we have taken so far) for $100*10^5$ states. The choices for each state are still "take the item" and "ignore the item", just like in usual knapsack.Maybe memory limit might be an issue, but I've never had problems with $10^7$ integer states.
•  » » » » 4 months ago, # ^ |   0 Can't seem to figure it out! If I include the element, I add its weight. And when I don't include the element, I don't add its weight. I have to take the minimum of both the choices. What will be the base case? When I reach the end of the array, I simply return 0. public static int solve2(int n, int w, int[][] arr, int maxV) { dp=new int[n+1][maxV+1]; for(int[] row:dp) { Arrays.fill(row, Integer.MAX_VALUE); } solveMemo(arr,0,0); for(int i=maxV;i>-1;--i) { if(dp[n-1][maxV]<=w) { return i; } } return 0; } static int[][] dp; public static int solveMemo(int[][] arr, int maxV,int idx) { if(idx==arr.length) { return 0; } if(dp[idx][maxV]!=Integer.MAX_VALUE) { return dp[idx][maxV]; } int include = arr[idx][0] + solveMemo(arr,maxV+arr[idx][1],idx+1); int not = solveMemo(arr,maxV,idx+1); return dp[idx][maxV]=Math.min(include, not); } But it doesn't seem to work! I also seem to understand while writing that it won't work. Where is it wrong and how do I make it work?
•  » » » » » 4 months ago, # ^ |   0 Recurrence looks ok to me :) But there are some issues in solve2 I think.int only holds up to $2^{31} - 1$, so it should overflow.Also in this if else if(dp[n-1][maxV]<=w) { return i; } Shouldn't it be more like this? if(dp[n-1][i]<=w) { return i; } 
•  » » » » » » 4 months ago, # ^ | ← Rev. 2 →   0 Fixed the solve2 method and overflow. Still doesn't give the correct answer. There's some issue in the recursion. Take case: 6 15 6 5 5 6 6 4 6 6 3 5 7 2 while calling recursion, at first, it goes on to include all elements, that is the first call, and then hits the base case from where it returns zero. Then the second call i.e. the not include case is called for the last element, from where it gets 0 from the base case. Now dp[n-1][maxV-valueOfLastElement]=0 because of the not include case. Isn't this wrong because it says to get value of [maxV-valueOfLastElement], we need only 0 weight. It didn't add up the weights of previously selected elements.
•  » » » » » » » 4 months ago, # ^ | ← Rev. 2 →   0 Yeah looks like we missed that part :)I guess we're both messing up the definitions a bit here. Usually we define dp(i, V) as "first i items, we are allowed to take V more values". So the answer should be dp(N, initialV), and the recurrence actually goes backwards from $N$ to $0$ (and likewise, from $maxV$ to $0$).So you can implement the same recurrence but backwards (i.e. "if I include the element, I decrease $maxV$"), and do for(int i = maxV;i>-1;--i) { if (solve(arr, N, i) <= w) { return i; } } 
•  » » 7 weeks ago, # ^ |   0 Thanks a lot for your hint!!
 » 11 months ago, # |   +2 You may consider iterating over sum of values rather than weights as per the constraints..So making the definition of dp[i] as minimum weight that can be taken with exactly i value..you can iterate on items and value(since constraints on value is low) My submission..https://atcoder.jp/contests/dp/submissions/10848260 ..Also you can refer to Youtube video by Errichto for the same
 » 11 months ago, # |   +8 MidoriFuse dark__forest thanks guys, that helped me
 » 11 months ago, # |   +5 An alternative is Branch and Bound algorithm. Don't simply look at its theoretical time complexity. It works wonders in practice.
 » 11 months ago, # | ← Rev. 2 →   0 You can try Branch and Bound to run faster Spoiler#include #include #include using namespace std; typedef long long ll; typedef vector vb; typedef vector vi; typedef vector vll; typedef vector vmb; /// Matrix boolean using vector typedef vector vmll; /// Matrix long long using vector const int INF = 1e9; /// Top-down knapsack using Branch and bound approach /// dp: calculate subproblem /// isDone: if subproblem is calculated before /// weight: weight of element /// value: value of element /// n: length of array /// i: index /// j: weight int tdkp01(vmll &dp, vmb &isDone, vi &weight, vi &value, int n, int i, int j) { if (j <= 0) /// Cant take more elements return 0; if (i == n) /// End of array return INF; if (isDone[i][j]) /// If solved before return dp[i][j]; isDone[i][j] = true; /// Solving Recursively return dp[i][j] = min(tdkp01(dp, isDone, weight, value, n, i + 1, j), tdkp01(dp, isDone, weight, value, n, i + 1, j - value[i]) + weight[i]); } /// Finding max sum where total weight <= limit int kp01(vmll &dp, vmb &isDone, vi &weight, vi &value, int n, int k, int vsum) { vsum++; while (vsum--) /// Finding from limit -> 0 for faster ending if (tdkp01(dp, isDone, weight, value, n, 0, vsum) <= k) return vsum; return 0; } int main() { int n, k; cin >> n >> k; vi weight(n), value(n); int vsum = 0; for (int i = 0; i < n; ++i) { cin >> weight[i] >> value[i]; vsum += value[i]; } vmll dp(n + 1); vmb isDone(n + 1); for (int i = 0; i <= n; ++i) { dp[i].assign(vsum, 0); isDone[i].assign(vsum, 0); } cout << kp01(dp, isDone, weight, value, n, k, vsum); return 0; } 
•  » » 11 months ago, # ^ | ← Rev. 3 →   +4 It should be noted that the Branch and Bound algorithm doesn't always run fast. It's runtime depends on the accuracy and speed of the cost-estimation function. The better the cost-estimation function, the faster your answer will attain the optimum value. In this case, the cost-estimation function is both good and fast, so Branch and Bound works.p.s. I don't think you understand how Branch and Bound works. It seems to me that your code uses DP instead of Branch and Bound.
•  » » » 11 months ago, # ^ |   0 Yes. Thank you for reminding me. Since I always adding Branch and Bound for faster in some specific problems, I forgot to notice that not at every time it work faster than normal approach
•  » » » 10 months ago, # ^ |   0 Yeah it is Top Down DP !!
 » 10 months ago, # |   0
 » 10 months ago, # |   0 This helped me.
 » 9 months ago, # | ← Rev. 2 →   0 How to solve Knapsack-2 in single-d array? Anyone? This is my code and i am really stuck. #include #include using namespace std; int main() { int n,w,gmax=0; cin>>n>>w; vector weights(n,0),values(n,0); for(int i=0;i>weights[i]>>values[i]; vector dp(100001); for(int i=0;i<=n;i++) { for(int j=0;j<100001;j++) { if(j==0) { dp[j]=0; continue; } if(i==0) { dp[j]=w+1; continue; } if(values[i-1]<=j) { dp[j]=min(weights[i-1]+dp[j-values[i-1]],dp[j]); } } } for(int j=0;j<100001;j++) { if(dp[j]<=w) gmax=max(gmax,j); } cout<
•  » » 9 months ago, # ^ | ← Rev. 2 →   0 Please put codes in spoilers. Your code is incorrect because You do not know if $dp[j-values[i-1]]$ has already used your current object and you may use it again. To fix that, you may iterate $j$ in reverse, so you know for a fact $dp[j-value[i-1]]$ is untouched.
•  » » » 9 months ago, # ^ |   0 Thanks a lot. It worked.
 » 9 months ago, # |   0 Refer to this solution in which I have explained Knapsnack 0-1 with all possible implementation 1. Recursive Memoisation 2. Tabulation: space O(n^2) 3. Tabulation: space O(n) 4. The special case when max capacity is large, done using tabulation.
•  » » 9 months ago, # ^ |   +1 Thanks
 » 9 months ago, # |   0 #include using namespace std; typedef long long ll; typedef pair pl; typedef pair pll; typedef vector vl; typedef vector vb; typedef vector vll; typedef vector vvl; typedef vector vvb; typedef vector vvll; typedef vector vpll; typedef vector vs; #define FOR(i,a,b) for(long long i=a;i=b;i--) #define F first #define S second #define pb push_back #define mp make_pair #define ub upper_bound #define lb lower_bound #define all(v) v.begin(),v.end() #define tc ll t;cin>>t;while(t--) #define io ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL) #define coutv(v) for(auto it: (v))cout<>v[i];} #define cinvg(v,n) (v).resize(n);FOR(i,0,(n)){cin>>v[i];} #define cin2d(v,n,m) vvll (v)(n,vll(m,0));FOR(i,0,n){FOR(j,0,m){cin>>v[i][j];}} #define cin2dg(v,n,m) (v).resize(n,vll(m));FOR(i,0,n){FOR(j,0,m){cin>>v[i][j];}} #define newl cout<<"\n" #define mod 1000000007 #define INF 1e18 int main() { io; ll n,W,V=0,ans=-1; cin>>n>>W; vll w(n+1),v(n+1); FOR(i,1,n+1) { cin>>w[i]>>v[i]; V+=v[i]; } vvll dp(n+1,vll(V+1,INF)); dp[0][0]=0; FOR(i,0,V+1) { dp[0][i]=0; } FOR(i,1,n+1) { FOR(j,1,V+1) { dp[i][j]=dp[i-1][j]; if(j-v[i]>=0){dp[i][j]=min(dp[i][j],dp[i-1][j-v[i]]+w[i]);} } } REV(i,V,1) { if(dp[n][i]<=W){ans=max(ans,i);} } cout<
 » 5 weeks ago, # | ← Rev. 4 →   0 In my dp[i][j] ,I stored the minimum weight required to calculate j value using i items Spoiler... n,w = map(int , input().split()) values = [] weights = [] ans= 0 totalvalues = 0 for i in range(n): a,b = map(int ,input().split()) weights.append(a) values.append(b) totalvalues += bdp = [ [0 for i in range(totalvalues+1)] for j in range(n+1) ] for i in range(totalvalues+1): dp[0][i] = float("inf") for i in range(n+1): dp[i][0] = 0for i in range(1,n+1): for j in range(1,totalvalues+1): if values[i-1] <= j : dp[i][j] = min(dp[i-1][j], dp[i-1][j-values[i-1]] + weights[i-1]) else : dp[i][j] = dp[i-1][j]for i in range(1,n+1): for j in range(1,totalvalues + 1): if dp[i][j] <= w: ans = jprint(ans)