hdu 2602 Bone Collector 01背包问题

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 66766    Accepted Submission(s): 27874

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 

#include<iostream>
#include<algorithm>
using namespace std;
#define  size 1010
//细节 题目要求最大值2的31次方 使用unsigned int
unsigned int volume[size], value[size];
unsigned int dp[size];
int main() {
	int test, n, v;
	cin >> test;
	while (test--) {
		while (cin >> n >> v) {
			memset(dp, 0, sizeof(dp));
			for (int i = 0; i < n; i++) {
				cin >> value[i];
			}
			for (int i = 0; i < n; i++) {
				cin >> volume[i];
			}
			//关键
			for (int i = 0; i < n; i++) {
				for (int j = v; j >= volume[i]; j--) {
					//背包问题的状态转移方程
					dp[j] = max(dp[j], dp[j - volume[i]] + value[i]);
				}
			}
			cout << dp[v] << endl;
		}
	}
}