HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 66304    Accepted Submission(s): 27660

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
1 5 10 1 2 3 4 5 5 4 3 2 1
 
Sample Output
14
 
Author
Teddy
 
Source
 
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吐槽:水题。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int T,n,m,w[1100],v[1100],f[1100];
int main(){
    scanf("%d",&T);
    while(T--){
        memset(f,0,sizeof(f));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&v[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&w[i]);
        for(int i=1;i<=n;i++)
            for(int j=m;j>=w[i];j--)
                f[j]=max(f[j],f[j-w[i]]+v[i]);
        printf("%d\n",f[m]);
    }
}

 

posted @ 2017-08-26 15:04  一蓑烟雨任生平  阅读(144)  评论(0编辑  收藏  举报