Bone Collector HDU 2602 DP 01 背包

Bone Collector

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 76372    Accepted Submission(s): 31644

 

 

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

01背包啥都没改，不解释了。。。。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;

int n, t, v,  dp[maxn], vaule[maxn], volume[maxn];

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