Hdu2602bone Collector Simple 0-1 backpack

Bone Collector

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 48618 Accepted Submission (s): 20269

Problem Description

Many years ago, in Teddy ' s hometown there is a man whowas called "Bone Collector". Collect varies of bones, such asdog ' s, cow ' s, also he went to the grave ...
The bone collector had a big bag with a volume of V, and along he trip of Collectingthere is a lot of bones, obviously , different bone have different value anddifferent volume, now given the each bone ' s value along his trip, can Youcalculat E 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 Twointeger N, V, (N <=, v <=) repre Senting the number of Bonesand the volume of his bag. And the second line contain N integers representingthe value of each bone. The third line contain N integers representing thevolume of each bone.

Output

One integer per line representing the maximum of Thetotal value (this number would be is less than 231).

Sample Input

1

5 10

1 2 3) 4 5

5 4 3) 2 1

Sample Output

14

Note: Enter the value of each bone first and then enter the volume of each bone.

State transition equation: Dp[j]=max (dp[j],dp[j-v[i]]+w[i].

#include <iostream>#include<string.h>using namespacestd;intn,v;intdp[1111];intw[1111],v[1111];intMain () {intT;  while(cin>>t) {         while(t--) {cin>>N>>V;  for(intI=1; i<=n;i++) {cin>>W[i]; }             for(intI=1; i<=n;i++) {cin>>V[i]; } memset (DP,0,sizeof(DP));  for(intI=1; i<=n;i++){                 for(intj=v;j>=v[i];j--) {//J if less than v[i] then v[i] must not be loaded into the bagDp[j]=max (dp[j],dp[j-v[i]]+W[i]); }} cout<<dp[V]<<Endl; }    }    return 0;}

Hdu2602bone Collector Simple 0-1 backpack