HDOJ_ACM_Bone Collector

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

 

Code

#include <stdio.h>
#include <algorithm>
using namespace std;
#define N 1000
struct Bone
{
    int value;
    int vol;
}bones[N + 5];
int f[N + 5][N + 5];
int main()
{
    int T, n, bagVol, i, j;
    scanf("%d", &T);
    while (T--)
    {
        //input and gain the radio which is the answer of value dividing volume
        scanf("%d %d", &n, &bagVol);
        for (i = 1; i <= n; i++)
            scanf("%d", &bones[i].value);
        for (i = 1; i <= n; i++)
            scanf("%d", &bones[i].vol);
        //handle
        for (i = 1; i <= n; i++)
            for (j = 0; j <= bagVol; j++)
                if (bones[i].vol <= j)
                    f[i][j] = max(f[i - 1][j], f[i - 1][j - bones[i].vol] + bones[i].value);
                else
                    f[i][j] = f[i - 1][j];
        //print
        printf("%d\n", f[n][bagVol]);
    }
    return 0;
}

Key Points

Firstly, you must realize it's a knapsack question. 

For those question, you should know the function f(n, v) = max(f(n - 1, v), f(n - 1, v - vol[n]) + value[n]).

f(n, v) means the greatest value of putting n things into knapsack whose volume is v.

vol[n] means the volume array.

value[n] means the value array.

If you don't choose the nth thing, it's f(n - 1, v), if you chooose the nth thing, it's f(n - 1, v - vol[n]) plus the nth thing's value.

then the bigger answer is the f(n, v).

Of course, you can using form to deepen understanding.

For example, if you input those number as follows.

1

2 5

3 4

2 3

	0	1	2	3	4	5
0	0	0	0	0	0	0
1	0	0	3	3	3	3
2	0	0	3	4	4	7

 

Secondly, I provide simpler program.

If you understand the first kind, I think it's easy for you. 

What I want to express is that the second circulation should begin with max, or you will be wrong. Because you put the same thing again.