HDU - 2602 - Bone Collector（01背包）

HDU - 2602 - Bone Collector

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 2 31).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14

题目链接

题目大意就是，给你一个背包装骨头，每块骨头都有自己的价值，并且还要占用一部分空间，要求在给定体积内怎么装才能获得价值最大。这个题目就是裸的01背包，一般的话，比较喜欢用一维数组，因为占空间少，写起来方便性也不输二维。下面直接上代码，套的模板：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define ll long long
using namespace std;
const int maxn =1e3 + 5;
int w[maxn], v[maxn];
ll dp[maxn];

int main()
{
    int t, n, V;
    scanf("%d", &t);
    while(t--)
    {
        memset(dp, 0, sizeof(dp));
        scanf("%d%d", &n, &V);
        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 = V; j >= w[i]; j--)
                dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
        printf("%lld\n", dp[V]);
    }
    return 0;
}