HDU-2602 Bone Collector 【01背包裸题】

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51478 Accepted Submission(s): 21677

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

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1. 题意：问背包最多能放下多少价值的骨头，每个骨头的价值和体积已知；
2. 思路：状态转移方程：f[i][v]=max(f[i-1][v],f[i-1][v-c[i]]+w[i])；
3. 失误：数据输反了；
4. 代码如下：

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#include
#include
#include
using namespace std;

const int  MAXN=1e3+11;
int dp[MAXN],c[MAXN],w[MAXN];

int main()
{
    int N,T,V,i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&N,&V);
        for(i=1;i<=N;++i)  scanf("%d",&w[i]);//输入反了 
        for(i=1;i<=N;++i)  scanf("%d",&c[i]);
        memset(dp,0,sizeof(dp));
        for(i=1;i<=N;++i)
        {
            for(j=V;j>=c[i];--j)
            {
                dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
            }
        }
        printf("%d\n",dp[V]);//写成N了 
    }
    return 0;
}