hdu 2602 Bone Collector

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题目描述:

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14889    Accepted Submission(s): 5911


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
 

Author
Teddy
 

Source
HDU 1st “Vegetable-Birds Cup” Programming Open Contest
 

Recommend
lcy

思路:

最原始的背包问题:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<set>#include<queue>#define inf 0xfffffff#define CLR(a,b) memset((a),(b),sizeof(a))using namespace std;int const nMax = 1100;int dp[nMax];int v[nMax],a[nMax];int n,V;int DP(){    CLR(dp,0);    for(int k=0;k<n;k++){        for(int i=V;i>=0;i--)if(i>=a[k]){            dp[i]=max(dp[i],max(dp[i-1],dp[i-a[k]]+v[k]));        }    }    return dp[V];}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&V);        for(int i=0;i<n;i++)scanf("%d",&v[i]);        for(int i=0;i<n;i++)scanf("%d",&a[i]);        printf("%d\n",DP());    }    return 0;}