HDU 2602 Bone Collector (01背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 45368    Accepted Submission(s): 18878

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 2 ³¹).

 

Sample Input

   
   
   
   

    
    
    
    1
5 10
1 2 3 4 5
5 4 3 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    14
   
   
   
   

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

刚刚学到DP  练练简单题：

题目就是和01背包完全一样  只不过物品改为了骨头！骨头有体积和价值！

有个坑就是可能有骨头体积为0，但依然有价值！！

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1000 + 10;
int dp[maxn][maxn],w[maxn],v[maxn],N,V;
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&N,&V);
		for (int i = 0; i < N; ++i)scanf("%d",&w[i]);
		for (int i = 0; i < N; ++i)scanf("%d",&v[i]);
		memset(dp,0,sizeof(dp));
		for (int i = 1; i <= N; ++i){	
			for (int j = 0; j <= V; ++j){
				if (j < v[i-1])dp[i][j] = dp[i-1][j];
				else dp[i][j] = max(dp[i-1][j],dp[i-1][j-v[i-1]]+w[i-1]);
			}
		}
		printf("%d\n",dp[N][V]);
	}
	return 0;
}