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 ?

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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.

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Output One integer per line representing the maximum of the total value (this number will be less than 2³¹).

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Sample Input 1 5 10 1 2 3 4 5 5 4 3 2 1

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Sample Output 14

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#include<iostream>
using namespace std;
int p[1001][1001];
int max(int a,int b){
if(a>=b)
return a;
else
return b;
}

int main(){
int a[1000],v[1000];
int t,n,c;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&c);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
scanf("%d",&v[i]);
memset(p,0,sizeof(p));

for(int i=1;i<=n;i++)
for(int j=0;j<=c;j++){
if(j>=v[i])
p[i][j]=max(p[i-1][j],p[i-1][j-v[i]]+a[i]);
else
p[i][j]=p[i-1][j];
}

/* for(int i=0;i<n;i++)
for(int j=c;j>=v[i];j--){
p[j]=max(p[j],p[j-v[i]]+a[i]);
}
*/
printf("%d\n",p[n][c]);

}

return 0;
}

以上兩種方法都可以接受，但注釋的方法更高效

解釋，可參考http://blog.sina.com.cn/s/blog_7e5541250100rtv5.html

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轉載于:https://www.cnblogs.com/hoojjack/p/4044181.html

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