Bone Collector

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

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

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

Recommend

表示都这个时候了才开始做dp  ，猛整他！

/*=============================================================================
#
#      Author: liangshu - cbam
#
#      QQ : 756029571
#
#      School : 哈尔滨理工大学
#
#      Last modified: 2015-08-28 19:18
#
#     Filename: B.cpp
#
#     Description:
#        The people who are crazy enough to think they can change the world, are the ones who do !
=============================================================================*/
#

#include<iostream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cctype>
#include<string>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;
struct A
{
    int val, v;
}E[1011];

int main(){
    int T;
    int dp[1011];
    cin>>T;
    int n, v;
    while(T--){
        scanf("%d%d",&n, &v);
        for(int i = 1; i <= n; i++){
            scanf("%d",&E[i].val);
        }
        for(int i = 1; i <= n; i++){
            scanf("%d",&E[i].v);
           }
        memset(dp, 0, sizeof(dp));
            for(int i = 1; i <= n; i++){
                for(int V = v; V >= E[i].v; V--){
                    dp[V] = max(dp[V], dp[V - E[i].v] + E[i].val);
                }

        }
    printf("%d\n",dp[v]);
    }
    return 0;
}

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