HDU2602 01背包入门 Bone Collector【01背包模板题】

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**

题意：容量为V的背包，尽可能装价值更多的石头，输出背包最大的价值
思路：01背包的经典套路，由石头1推到石头n，第i个石头下背包容量V的最大价值由i-1递推而来

#include<stdio.h>
#include<string.h>
#define N 1000
int num[N+10],w[N+10],v[N+10];
int main()
{
    int i, j;
    int t, n, V;
    scanf("%d",&t);
    while( t --)
    {
        memset(num,0,sizeof(num));
        memset(w,0,sizeof(w));
        memset(v,0,sizeof(v));
        scanf("%d%d",&n,&V);
        for(i = 1; i <= n; i ++)
            scanf("%d",&w[i]);
        for( i = 1; i <= n; i ++)
            scanf("%d",&v[i]);
        for( i = 1; i <= n; i ++)
            for( j = V; j >=v[i];j --)/*j代表背包的剩余容量*/
            {
                num[j] = max(num[j],num[j-v[i]]+w[i]);
                /*num[j]表示不把第i个石头放入背包的价值*/
                /*num[j-v[i]]+w[i]表示把第i个石头放入背包后的价值*/
                /*比较二者的最优解存入num[j]*/
            }
        printf("%d\n",num[V]);
    }
    return 0;
}