【HDU

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

分析 01背包的模板
代码

#include<bits/stdc++.h>#define LL long longusing namespace std;const int MAXN = 1e3+10;const int MAXM = 1e5;int dp[MAXN]; int cost[MAXN],val[MAXN];int main(){    int t;cin>>t;    while(t--){        int n,v;scanf("%d%d",&n,&v);        for(int i=1;i<=n;i++)            scanf("%d",&val[i]);        for(int i=1;i<=n;i++)            scanf("%d",&cost[i]);        memset(dp,0,sizeof(dp));//  因为题目没有要求要恰好装满        for(int i=1;i<=n;i++){            for(int j=v;j>=cost[i];j--)                dp[j]=max(dp[j],dp[j-cost[i]]+val[i]);        }         printf("%d\n",dp[v])    ;    }     return 0;}