Bone Collector 01背包

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 ?

 

输入

 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.

输出

 One integer per line representing the maximum of the total value (this number will be less than 231).

 

示例输入

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

示例输出

14

#include<stdio.h>
#include<string.h>
int c[1100][1100],p[1100],w[1100];
//c[i][j]表示前i件物品恰好放入容积为j的背包获得的价值
int knaspack(int m, int n)
{
    int i, j;
    memset(c,0,sizeof(c));
    for(i = 1; i < m+1; i++)
        for(j = 1; j< n+1; j++)
    {
//如果第i件物品的体积小于容积j时，可包含两种情况：
（1）物品i放入背包，前i-1件物品的体积为j-w[i],所以c[i][j]=c[i-1][j-w[i]]+p[i];
 (2)物品i不放入背包中，则c[i][j] = c[i-1][j];
        if(w[i] <= j)
        {
            if(c[i-1][j-w[i]]+p[i] > c[i-1][j])
                c[i][j] = c[i-1][j-w[i]] + p[i];
            else c[i][j] = c[i-1][j];
        }
        else c[i][j] = c[i-1][j];
    }
    return c[m][n];
}
int main ()
{
    int t,i,m,n;
    scanf("%d",&t);


    while(t--)
    {
        scanf("%d %d",&m, &n);
//有m件物品，容器最大容积为n
        for(i = 1; i <= m; i++)
            scanf("%d",&p[i]);
//物品的价值
        for(i = 1; i<= m; i++)
            scanf("%d",&w[i]);
//物品的体积。
        printf("%d\n",knaspack(m,n));
    }
    return 0;
}

 也可以用一位数组

#include<stdio.h>
#include<string.h>
int p[1100],w[1100],dp[1100];
int main ()
{
    int n,m,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        for(int i = 0; i < n; i++)
            scanf("%d",&p[i]);
        for(int i = 0; i < n; i++)
            scanf("%d",&w[i]);
        memset(dp,0,sizeof(dp));
        for(int i = 0; i < n; i++)
            for(int j = m;j >= w[i]; j--)//体积按倒序
                if(dp[j-w[i]]+p[i] > dp[j])
                    dp[j] = dp[j-w[i]]+p[i];
        printf("%d\n",dp[m]);
    }
    return 0;

}