HDU 2602 Bone Collector

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 18705    Accepted Submission(s): 7386

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

背包第一题。

#include<stdio.h>#include<string.h>int dp[1007];struct sa{    int W,V;}data[1007];int max(int a,int b){    return a>b?a:b;}int main(){    int t,i,j;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        int n,v;        scanf("%d%d",&n,&v);        for(i=1;i<=n;i++)        scanf("%d",&data[i].W);        for(j=1;j<=n;j++)        scanf("%d",&data[j].V);        for(i=1;i<=n;i++)        for(j=v;j>=data[i].V;j--)        {            dp[j]=max(dp[j],dp[j-data[i].V]+data[i].W);        }        printf("%d\n",dp[v]);    }    return 0;}