HDU Problem 2062 Bone Collector【01背包】

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51847    Accepted Submission(s): 21829

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

 

 

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状态转移方程：dp[i][j] = max{dp[i+1][j], dp[i+1][j-vol[i]]+val[i]};其中i表示第i个物品，j表示背包当前的容量，而数组中所存放的元素为在当前状态下的最大价值。通过下面的每个状态的最大值向上推。因此，最终的最大状态应该存放在dp[0][v]中。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1000 + 10;
const int INF = 0x3f3f3f3f;
int  n, v, val[MAXN], vol[MAXN], dp[MAXN][MAXN];
int main() {
    int t; scanf("%d", &t);
    while (t--) {
        scanf("%d %d", &n, &v);
        for (int i = 0; i < n; i++) {
            scanf("%d", &val[i]);
        }
        for (int i = 0; i < n; i++) {
            scanf("%d", &vol[i]);
        }
        memset(dp, 0, sizeof(dp));
        for (int i = n - 1; i >= 0; i--) {
            for (int j = 0; j <= v; j++) {
                if (j < vol[i]) dp[i][j] = dp[i + 1][j];
                else dp[i][j] = max(dp[i+1][j], dp[i+1][j-vol[i]]+val[i]);
            }
        }
        printf("%d\n", dp[0][v]);
    }
    return 0;
}