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  • HDU 2602 Bone Collector 0/1背包

    题目链接:HDU 2602 Bone Collector

    Bone Collector

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 28903    Accepted Submission(s): 11789


    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
     

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    经典0/1背包问题。

    有N件物品和一个容量为V的背包,第i件物品的体积为v[i],价值为w[i],求解将哪些物品装入背包才干使总价值最大。

    这是最基础的背包问题,每种物品仅仅有一件。且仅仅有两种状态:放与不放。

    用子问题定义状态:dp[i][v]表示前i件物品恰好放入一个容量为m的包中可获得的最大价值。

    状态转移方程:dp[i][m] = max(dp[i-1][m], dp[i-1][m-v[i]]+w[i]);

    可转化为一维情况:dp[m] = max(dp[m], dp[m-v[i]]+w[i]);

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    
    int n, v, dp[1010], value[1010], volume[1010];
    int main()
    {
        int t;
        scanf("%d", &t);
        while(t--)
        {
            scanf("%d%d", &n, &v);
            for(int i = 1; i <= n; i++)
                scanf("%d", &value[i]);
            for(int i = 1; i <= n; i++)
                scanf("%d", &volume[i]);
            memset(dp, 0, sizeof(dp));
            for(int i = 1; i <= n; i++)
                for(int j = v; j >= volume[i]; j--)
                    dp[j] = max(dp[j], dp[j-volume[i]]+value[i]);
            printf("%d
    ", dp[v]);
        }
    
        return 0;
    }
    






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  • 原文地址:https://www.cnblogs.com/claireyuancy/p/7395512.html
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