Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1760
Solution idea:
This is a classical knapsack problem. Set a correct base case is the most challenging part of this problem. If we read and think carefully about this problem we can see that if the expense exceeds 2000$ then 200$ will be refunded so it will be strictly greater than 2000. Now we can consider 3 things as base case. Those are here-
Let W is expense at current time and C = capacity or Total money.
- When W > Capacity and W-Capacity > 200 then we can’t buy anything else and the current product so we will return a large negative value to cancel this solution.
- When Capacity < 1800 and Capacity < W then it's clear we can't buy anything and current product so we will cancel this.
- And Last thing is when we are taking the last product then if W > Capacity and W <= 2000 then we can take these so we will return a Zero. Otherwise we will cancel this one by returning a large negative value.
/* +-+ +-+ +-+ |R| |.| |S| +-+ +-+ +-+ */ #include <bits/stdc++.h> #define pii pair <int,int> #define sc scanf #define pf printf #define Pi 2*acos(0.0) #define ms(a,b) memset(a, b, sizeof(a)) #define pb(a) push_back(a) #define MP make_pair #define db double #define ll long long #define EPS 10E-10 #define ff first #define ss second #define sqr(x) (x)*(x) #define D(x) cout<<#x " = "<<(x)<<endl #define VI vector <int> #define DBG pf("Hi\n") #define MOD 100007 #define MAX 10000 #define CIN ios_base::sync_with_stdio(0); cin.tie(0) #define SZ(a) (int)a.size() #define sf(a) scanf("%d",&a) #define sfl(a) scanf("%lld",&a) #define sff(a,b) scanf("%d %d",&a,&b) #define sffl(a,b) scanf("%lld %lld",&a,&b) #define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c) #define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define loop(i,n) for(int i=0;i<n;i++) #define REP(i,a,b) for(int i=a;i<b;i++) #define TEST_CASE(t) for(int z=1;z<=t;z++) #define PRINT_CASE printf("Case %d: ",z) #define all(a) a.begin(),a.end() #define intlim 2147483648 #define inf 1000000 #define ull unsigned long long using namespace std; /*----------------------Graph Moves----------------*/ //const int fx[]={+1,-1,+0,+0}; //const int fy[]={+0,+0,+1,-1}; //const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move //const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move //const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move //const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move /*------------------------------------------------*/ /*-----------------------Bitmask------------------*/ //int Set(int N,int pos){return N=N | (1<<pos);} //int reset(int N,int pos){return N= N & ~(1<<pos);} //bool check(int N,int pos){return (bool)(N & (1<<pos));} /*------------------------------------------------*/ int capacity, n; int z; int dp[110][10300]; int weight[110],profit[110]; int knapsack(int idx, int w) { if(w>capacity && w-capacity>200) return -100000; if(capacity<1800 && w>capacity) return -100000; if(idx==n) { if(w>capacity && w<=2000) return -100000; return 0; } if(dp[idx][w]!=-1) return dp[idx][w]; return dp[idx][w]=max(profit[idx]+knapsack(idx+1,w+weight[idx]),knapsack(idx+1,w)); } int main() { ///freopen("in.txt","r",stdin); ///freopen("out.txt","w",stdout); CIN; while(cin>>capacity>>n) { z++; loop(i,n) cin>>weight[i]>>profit[i]; ms(dp,-1); cout<<knapsack(0,0)<<endl; } return 0; }
Here is another solution 🙂
/* +-+ +-+ +-+ |R| |.| |S| +-+ +-+ +-+ */ #include <bits/stdc++.h> #define pii pair <int,int> #define sc scanf #define pf printf #define Pi 2*acos(0.0) #define ms(a,b) memset(a, b, sizeof(a)) #define pb(a) push_back(a) #define MP make_pair #define db double #define ll long long #define EPS 10E-10 #define ff first #define ss second #define sqr(x) (x)*(x) #define D(x) cout<<#x " = "<<(x)<<endl #define VI vector <int> #define DBG pf("Hi\n") #define MOD 100007 #define MAX 10000 #define CIN ios_base::sync_with_stdio(0); cin.tie(0) #define SZ(a) (int)a.size() #define sf(a) scanf("%d",&a) #define sfl(a) scanf("%lld",&a) #define sff(a,b) scanf("%d %d",&a,&b) #define sffl(a,b) scanf("%lld %lld",&a,&b) #define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c) #define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define loop(i,n) for(int i=0;i<n;i++) #define REP(i,a,b) for(int i=a;i<b;i++) #define TEST_CASE(t) for(int z=1;z<=t;z++) #define PRINT_CASE printf("Case %d: ",z) #define all(a) a.begin(),a.end() #define intlim 2147483648 #define inf 1000000 #define ull unsigned long long using namespace std; /*----------------------Graph Moves----------------*/ //const int fx[]={+1,-1,+0,+0}; //const int fy[]={+0,+0,+1,-1}; //const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move //const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move //const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move //const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move /*------------------------------------------------*/ /*-----------------------Bitmask------------------*/ //int Set(int N,int pos){return N=N | (1<<pos);} //int reset(int N,int pos){return N= N & ~(1<<pos);} //bool check(int N,int pos){return (bool)(N & (1<<pos));} /*------------------------------------------------*/ int capacity, n; pii input[110]; int z; int dp[110][10210]; int cases[110][10210]; bool test=0; int knapsack(int idx, int w) { if(w==capacity) { return 0; } if(idx>=n) return 0; int &ret=dp[idx][w]; int &cas=cases[idx][w]; if(cas==z) return ret; cas=z; int p=0,q=0; if(w+input[idx].ff<=capacity) p=input[idx].ss+knapsack(idx+1,w+input[idx].ff); else if(w+input[idx].ff>capacity) { if((w+input[idx].ff>2000) && (w+input[idx].ff<=capacity+200) && !test) { test=1; capacity+=200; p=input[idx].ss+knapsack(idx+1,w+input[idx].ff); test=0; capacity-=200; } } q=knapsack(idx+1,w); return ret=max(p,q); } int main() { ///freopen("in.txt","r",stdin); ///freopen("out.txt","w",stdout); while(cin>>capacity>>n) { z++; loop(i,n) cin>>input[i].ff>>input[i].ss; test=0; sort(input,input+n); cout<<knapsack(0,0)<<endl; } return 0; }