hdu 2602 Bone Collector(01背包) - c++编程基础

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24175 Accepted Submission(s): 9809

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 2 ³¹).
Sample Input

Sample Output

最基础的01背包。给n个物品的价值和花费以及背包的容量，每个物品只有一件。问背包所放物品的最大价值。 f[ v ] = max ( f[ v ], f[ v - c[i] ] + w[i] )。

#include 
  
   
#include 
   
     #include 
    
      using namespace std; int f[1010],c[1010],w[1010]; int main() { int test,n,m; scanf("%d",&test); while(test--) { scanf("%d %d",&n,&m); for(int i = 1; i <= n; i++) scanf("%d",&w[i]); for(int i = 1; i <= n; i++) scanf("%d",&c[i]); memset(f,0,sizeof(f)); for(int i = 1; i <= n; i++) { for(int j = m; j >= c[i]; j--)//体积逆序 { f[j] = max(f[j], f[j-c[i]]+w[i]); } } printf("%d\n",f[m]); } return 0; }