HDU

先上题目：

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23533    Accepted Submission(s): 9535

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

1

5 10

1 2 3 4 5

5 4 3 2 1

 

Sample Output

14

 

　　这一题说白了就是一条01背包的模板，只是因为物品数目最多有1000种，所以要么使用滚动数组，要么只用一维即可。滚动数组比较好实现，所以这里我用一维数组实现，但这里需要注意的是，如果是使用一位数组实现，因为原本的状态转移方程是：

　　　　　　　　 max{dp[i-1][j],dp[i-1][j-c[i]]+v[i]} 　　c[i]<=j

　　　   dp[i][j]=dp[i-1][j]　　　　　　　　　　　　　　　 c[i]>j

　　　　　　　　 0　　　　　　　　　　　　　　　　　　　 i==0 || j==0

　　消去一维以后因为需要用到j-c[i]的数据，所以外面的循环是从1~n，里面的循环是从C~0(n是物品的数目，C是背包的总体积)，只有这样做j-c[i]的数据才不会在被使用之前被覆盖。

 

上代码：

 

#include <cstdio>
#include <cstring>
#define max(x,y) (x > y ? x : y)
#define MAX 1001
using namespace std;

int dp[MAX],n,s;
int v[MAX],c[MAX];

int main()
{
    int t;
    //freopen("data.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&s);
        for(int i=1;i<=n;i++){
            scanf("%d",&v[i]);
        }
        for(int i=1;i<=n;i++){
            scanf("%d",&c[i]);
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            for(int j=s;j>=0;j--){
                if(c[i]<=j) dp[j]=max(dp[j],(dp[j-c[i]]+v[i]));
                //printf("%d ",dp[j]);
            }
            //printf("
");
        }
        printf("%d
",dp[s]);
    }
    return 0;
}