基础-DP
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
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The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
Input
The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Output
One integer per line representing the maximum of the total value (this number will be less than 2 31).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
#include<iostream> using namespace std; #include<math.h> #include<string.h> #define maxn 100000 long long d[maxn+5]; int max(int a,int b) { if(a>b) return a; else return b; } int main() { int t,n,m,i,j; int V[1005],N[1005]; cin>>t; while(t--) { memset(d,0,sizeof(d)); cin>>n>>m; for(i=0;i<n;i++) cin>>V[i]; for(j=0;j<n;j++) cin>>N[j]; for(i=0;i<n;i++) for(j=m;j>=N[i];j--) { d[j]=max(d[j],d[j-N[i]]+V[i]); } cout<<d[m]<<endl; } return 0; }