HDU - 2602 Bone Collector

Bone Collector

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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 2   ³¹).

 

Sample Input

    

     1
5 10
1 2 3 4 5
5 4 3 2 1 
    

 

Sample Output

    

     14 
    

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

分析：

裸0-1背包。

经典的01背包题，给出了石头的数量与背包的容量，然后分别给出每个石头的容量与价值，要求最优解。

ac代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct Node
{
    int h;
    int v;
} node[1005];

int main()
{
    int t,n,m,l;
    int dp[1005];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        int i;
        for(i = 1; i<=n; i++)
            scanf("%d",&node[i].h);
        for(i = 1; i<=n; i++)
            scanf("%d",&node[i].v);
        memset(dp,0,sizeof(dp));
        for(i = 1; i<=n; i++)
        {
            for(l = m; l>=node[i].v; l--)
                dp[l] = max(dp[l],dp[l-node[i].v]+node[i].h);
        }
        printf("%d\n",dp[m]);
    }

    return 0;
}