HDOJ 2602 Bone Collector(01背包最基本模板题)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 52215    Accepted Submission(s): 22001


[align=left]Problem Description[/align]
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 ?



 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
One integer per line representing the maximum of the total value (this number will be less than 231).
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

14

 

[align=left]Author[/align]
Teddy
 

[align=left]Source[/align]
HDU 1st “Vegetable-Birds Cup” Programming Open Contest 

思路：

这道题是最基本的01背包模板题，直接套模板即可。

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<map>
#include<queue>
#include<cmath>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
#define LL long long
#define inf 1<<29
using namespace std;
const int N=1005;
int n,m,a,b,c;
int cost
;
int v
;
int dp

;
int main()
{
int t,w;
scanf("%d",&t) ;
while(t--){
scanf("%d",&n);
scanf("%d",&w);
int sum=0;
for(int i=1;i<=n;i++) scanf("%d",&v[i]) ;
for(int i=1;i<=n;i++) scanf("%d",&cost[i]) ;
memset(dp,0,sizeof(dp));
dp[0][0]=0;
for(int i=1;i<=n;i++)
for(int j=w;j>=0;j--)
{
if(j>=cost[i])dp[i][j]=max(dp[i-1][j],dp[i-1][j-cost[i]]+v[i]);
else dp[i][j]=dp[i-1][j];
}
printf("%d\n",dp
[w]);
}
return 0;
}