hdu 2602 Bone Collector

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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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14就是01背包真接上代碼
#include<stdio.h>int c[3500],w[3500],f[13000];int main(){int i,j,k,n,v,m,t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=0;i<=m;i++)f[i]=0;for(i=0;i<n;i++)scanf("%d",w+i);for(i=0;i<n;i++)scanf("%d",c+i);for (i = 0; i < n; ++i){for (v = m; v >= c[i]; --v) //注意這裡的Ｖ是從大到小c[i]可最佳化為bound,bound = max {V - sum c[i,...n],c[i]}{f[v] = (f[v] > f[v - c[i]] + w[i]?f[v] : f[v - c[i]] + w[i]);}}printf("%d\n",f[m]);}return 0;}