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

 1 #include<stdio.h>

 2 #include<string.h>

 3 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
 4 int knaspack(int m, int n)

 5 {

 6     int i, j;

 7     memset(c,0,sizeof(c));

 8     for(i = 1; i < m+1; i++)

 9         for(j = 1; j< n+1; j++)

10     {

//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];
11         if(w[i] <= j)

12         {

13             if(c[i-1][j-w[i]]+p[i] > c[i-1][j])

14                 c[i][j] = c[i-1][j-w[i]] + p[i];

15             else c[i][j] = c[i-1][j];

16         }

17         else c[i][j] = c[i-1][j];

18     }

19     return c[m][n];

20 }

21 int main ()

22 {

23     int t,i,m,n;

24     scanf("%d",&t);

25 

26 

27     while(t--)

28     {

29         scanf("%d %d",&m, &n);

//有m件物品，容器最大容积为n
30         for(i = 1; i <= m; i++)

31             scanf("%d",&p[i]);

//物品的价值
32         for(i = 1; i<= m; i++)

33             scanf("%d",&w[i]);

//物品的体积。
34         printf("%d\n",knaspack(m,n));

35     }

36     return 0;

37 }

 也可以用一位数组

 1 #include<stdio.h>

 2 #include<string.h>

 3 int p[1100],w[1100],dp[1100];

 4 int main ()

 5 {

 6     int n,m,t;

 7     scanf("%d",&t);

 8     while(t--)

 9     {

10         scanf("%d %d",&n,&m);

11         for(int i = 0; i < n; i++)

12             scanf("%d",&p[i]);

13         for(int i = 0; i < n; i++)

14             scanf("%d",&w[i]);

15         memset(dp,0,sizeof(dp));

16         for(int i = 0; i < n; i++)

17             for(int j = m;j >= w[i]; j--)//体积按倒序

18                 if(dp[j-w[i]]+p[i] > dp[j])

19                     dp[j] = dp[j-w[i]]+p[i];

20         printf("%d\n",dp[m]);

21     }

22     return 0;

23 

24 }