Acm1881 01 backpack problem Application

01 dynamic planning and Application of knapsack problems

Acm1881Graduation BG

Think of the time limit that must be left as the size of the backpack, first sort them in ascending order, and then start with the time limit for each instance in the sorted array (the size of the backpack) to its extended time for traversal;

 1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 using namespace std; 5 struct BG 6 { 7     int h,t,l; 8     friend bool operator<(BG a,BG b)     9     {10         return a.l<b.l;11     }12 };13 BG *bg;14 int main()15 {16     int n,m;17     int value[3500];18     while(cin>>n&&n>=0)19     {20         m=0;21         if(n==0)22         {23             cout<<0<<endl;24             continue;25         }26         bg=new BG[n];27         memset(value,0,sizeof(value));28         for(int i=0;i<n;i++)29         {30             cin>>bg[i].h>>bg[i].t>>bg[i].l;31             m=(bg[i].l>m)?bg[i].l:m;32         }33         sort(bg,bg+n);34         for(int i=0;i<n;i++)35         {36             for(int j=bg[i].l;j>=bg[i].t;j--)37             {38                 value[j]=max(value[j],value[j-bg[i].t]+bg[i].h);39             }40         }41         int ans=0;42         for(int i=0;i<=m;i++)43         {44             if(value[i]>ans)ans=value[i];45         }46         cout<<ans<<endl;47     }48     return 0;49 }