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HDU - 2602 Bone Collector

时间:2020-02-16 23:26:01      阅读:59      评论:0      收藏:0      [点我收藏+]
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 ?技术分享图片

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 2 31).Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14


#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
const int maxn = 1010;
const int inf = 0x3f3f3f3f;
int n;
int v;
int f[maxn];
int c[maxn];
int w[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d", &n, &v);
        memset(f, 0, sizeof f);
        for (int i = 1; i <= n; i++) cin >> w[i];
        for (int i = 1; i <= n; i++) cin >> c[i];
        for (int i = 1; i <= n; ++i)
          for (int j = v; j >= c[i]; --j)
            f[j] = max(f[j], f[j - c[i]] + w[i]);
        printf("%d\n", f[v]);
    }
    return 0;
}

 

 

HDU - 2602 Bone Collector

原文:https://www.cnblogs.com/-113/p/12319166.html

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