01-背包问题---Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35547    Accepted Submission(s): 14631

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 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 
未进行优化的代码：

#include <iostream>#include <cstring>using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */int max(int a,int b){return (a>b)?a:b;}int val[1005],vol[1005];int v[1005][1005];int main(int argc, char *argv[]) {int t;int n,vv;cin>>t;while(t--){cin>>n>>vv;for(int i=1;i<=n;i++)cin>>val[i];for(int i=1;i<=n;i++)cin>>vol[i];for(int i=0;i<=n;i++)v[i][0]=0;for(int j=0;j<=vv;j++)v[0][j]=0;memset(v,0,sizeof(v));int i,j;for(i=1;i<=n;i++){for(j=0;j<=vv;j++){if(vol[i]<=j){v[i][j]=max(v[i-1][j],v[i-1][j-vol[i]]+val[i]);}elsev[i][j]=v[i-1][j];}}cout<<v[n][vv]<<endl;}return 0;}

优化后：

#include <iostream>#include <cstring>using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */int max(int a,int b){return (a>b)?a:b;}int val[1005],vol[1005];int v[1005];int main(int argc, char *argv[]) {int t;int n,vv;cin>>t;while(t--){cin>>n>>vv;for(int i=1;i<=n;i++)cin>>val[i];for(int i=1;i<=n;i++)cin>>vol[i];for(int i=0;i<=vv;i++)v[i]=0;memset(v,0,sizeof(v));int i,j;for(i=1;i<=n;i++){for(j=vv;j>=0;j--){if(vol[i]<=j){v[j]=max(v[j],v[j-vol[i]]+val[i]);}}}cout<<v[vv]<<endl;}return 0;}