hdoj - 2602 Bone Collector
Problem Description
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 1
Sample Output
14
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int a[1005],b[1005],dp[1005]; int main() { int t,n,v; scanf("%d",&t); for(int k=0;k<t;k++) { memset(dp,0,sizeof(dp)); scanf("%d %d",&n,&v); for(int i = 0;i<n;i++) { scanf("%d",&a[i]); } for(int i = 0;i<n;i++) { scanf("%d",&b[i]); } for(int i = 0;i<n;i++) { for(int j = v;j>=b[i];j--) { dp[j] = max(dp[j],dp[j-b[i]]+a[i]); //关键代码 } } printf("%d\n",dp[v]); } return 0; }