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hdu 2602 Bone Collector (01背包)
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27002 Accepted Submission(s): 10938
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 ?
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.
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
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
Author
Teddy
Source
HDU 1st “Vegetable-Birds Cup” Programming Open Contest
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01背包模板题:
1 //15MS 240K 545 B C++ 2 #include<stdio.h> 3 #include<string.h> 4 int dp[1005]; 5 int v[1005],w[1005]; 6 int Max(int a,int b) 7 { 8 return a>b?a:b; 9 } 10 int main(void) 11 { 12 int t,n,m; 13 scanf("%d",&t); 14 while(t--) 15 { 16 scanf("%d%d",&n,&m); 17 memset(dp,0,sizeof(dp)); 18 for(int i=1;i<=n;i++) scanf("%d",&v[i]); 19 for(int i=1;i<=n;i++) scanf("%d",&w[i]); 20 for(int i=1;i<=n;i++) 21 for(int j=m;j>=w[i];j--) 22 dp[j]=Max(dp[j],dp[j-w[i]]+v[i]); 23 printf("%d\n",dp[m]); 24 } 25 return 0; 26 }
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