【模板--完全背包】HDU--2602 Bone Collector

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 ?
【模板--完全背包】HDU--2602  Bone Collector
 
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
1
5 10
1 2 3 4 5
5 4 3 2 1
 
Sample Output
14
 #include<cstdio>
#include<algorithm>
using namespace std;
long long dp[+];
int w[],p[];
int main()
{
int t,N,W;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&N,&W);
for(int i=;i<=N;i++) //
scanf("%d",&p[i]);
for(int i=;i<=N;i++) //
scanf("%d",&w[i]);
for(int i=;i<=W;i++)
dp[i]=;
for(int i=;i<=N;i++) // 这几个地方i开头值注意一样 错了几次..
{
for(int j=W;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+p[i]); //核心代码
}
}
printf("%lld\n",dp[W]); } return ;
}
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