HDU 2602--Bone Collector【01背包】

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 34467    Accepted Submission(s): 14191

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 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

如果要求恰好装满背包，那么在初始化时除了dp[0]为0其它dp[1..V]均设为-∞，这样就可以保证最终得到的f[N]是一种恰好装满背包的最优解。
如果并没有要求必须把背包装满，而是只希望价格尽量大，初始化时应该将dp[0..V]全部设为0。
为什么呢？可以这样理解：初始化的f数组事实上就是在没有任何物品可以放入背包时的合法状态。如果要求背包恰好装满，
那么此时只有容量为0的背包可能被价值为0的nothing“恰好装满”，其它容量的背包均没有合法的解，属于未定义的状态，
它们的值就都应该是-∞了。如果背包并非必须被装满，那么任何容量的背包都有一个合法解“什么都不装”，这个解的价值为0，
所以初始时状态的值也就全部为0了。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define max(a,b) ((a)>(b)?(a):(b))#define maxn 1100int val[maxn],vol[maxn],dp[maxn];int main (){    int t,i,j,n,v;    scanf("%d",&t);    while(t--){        memset(dp,0,sizeof(dp));        scanf("%d%d",&n,&v);        for(i=1;i<=n;++i)            scanf("%d",&val[i]);        for(i=1;i<=n;++i)            scanf("%d",&vol[i]);        for(i=1;i<=n;++i)            for(j=v;j-vol[i]>=0;j--)                dp[j]=max(dp[j-vol[i]]+val[i],dp[j]);        printf("%d\n",dp[v]);    }    return 0;}