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 ?
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 231).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1Sample Output
14
动态规划的学习,但是我还是搞不太懂,如何推公式
#include<stdio.h> #include<string.h> #include<algorithm> #include<cmath> #include<iostream> using namespace std; struct BONE { int val; int vol; }bone[1011]; int N,V; int dp[1011][1011]; int ans() { memset(dp,0,sizeof(dp)); for(int i=1;i<=N;i++) { for(int j=0;j<=V;j++) { if(bone[i].vol>j) { dp[i][j]=dp[i-1][j]; } else { dp[i][j]=max(dp[i-1][j],dp[i-1][j-bone[i].vol]+bone[i].val); } } } return dp[N][V]; } int main() { int t; cin>>t; while(t--) { cin>>N>>V; for(int i=1;i<=N;i++) { cin>>bone[i].val; } for(int i=1;i<=N;i++) { cin>>bone[i].vol; } cout<<ans()<<endl; } return 0; }