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HDU--dp练习--1007--Bone Collector

热度:21   发布时间:2024-01-12 14:33:55.0

题目:

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 2 31).

 

Sample Input
5 10 
1 2 3 4 5 
5 4 3 2 1

 

Sample Output
14
题目大意:

有一个收集骨头的人,现在有n种骨头,背包总容量为v,现在给出n和v,以及ni种骨头的价值以及体积,求所能收集到的最大值。

思路:

01背包问题。

源代码:

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int N;
    int n,m,i,t;
    cin >> N;
    int V[10005],value[10005];
    int dp[10005];
    while (N--)
    {
        cin >> n >> m;
        for (i = 1; i <= n; i++)
            cin >> value[i];
        for (i = 1; i <= n; i++)
            cin >> V[i];
        memset(dp,0,sizeof(dp));
        for (i = 1; i <= n; i++)
            for (t = m; t >= V[i] ; t--)
            {
                if (dp[t] < dp[t - V[i]]+value[i])
                    dp[t] = dp[t - V[i]]+value[i];
            }
        cout << dp[m] << endl;
    }
    return 0;
}

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