HDU2602 Bone Collector 【01背包】[通俗易懂]

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Bone Collector

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

Total Submission(s): 28365 Accepted Submission(s): 11562

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 ?

HDU2602 Bone Collector 【01背包】[通俗易懂]

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

01背包入门题。

#include <stdio.h>
#include <string.h>
#define maxn 1002

int dp[maxn], w[maxn], v[maxn];

int main()
{
    int t, n, val, i, j;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &n, &val);
        for(i = 1; i <= n; ++i) scanf("%d", v + i);
        for(i = 1; i <= n; ++i) scanf("%d", w + i);
        memset(dp, 0, sizeof(dp));
        for(i = 1; i <= n; ++i){
            for(j = val; j >= w[i]; --j){
                if(dp[j] < dp[j-w[i]] + v[i]) 
                    dp[j] = dp[j-w[i]] + v[i];
            }
        }
        printf("%d\n", dp[val]);
    }
    return 0;
}

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