【HDU 2602】Bone Collector(01背包)
2016-07-25 21:02
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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 ?
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.
Output
One integer per line representing the maximum of the total value (this number will be less than 231).Sample Input
15 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14题目大意
背包问题,给你两个数N,V分别表示骨头的个数和背包的容量,接下来一行表示每个骨头的价值,在接下来的一行表示每个骨头的重量。求最大能被价值为多少的骨头。思路
01背包裸体代码
#include <iostream> #include <cstdio> #include <cstring> #includ 4000 e <algorithm> using namespace std; const int maxn=1000+5; int w[maxn],v[maxn],dp[maxn],t,n,cost; int main() { scanf("%d",&t); while(t--) { memset(dp,0,sizeof(dp)); scanf("%d %d",&n,&cost); for(int i=1;i<=n;i++) { scanf("%d",&v[i]); } for(int i=1;i<=n;i++) { scanf("%d",&w[i]); } for(int i=1;i<=n;i++) { for(int j=cost;j>=w[i];j--) { dp[j]=max(dp[j],dp[j-w[i]]+v[i]); } } printf("%d\n",dp[cost]); } return 0; }
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