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  • HDU 2602 Bone Collector DP(01背包)

    Bone Collector
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
     

    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

    1
    5 10
    1 2 3 4 5
    5 4 3 2 1

    Sample Output

    14

    很基础的一道dp
    熟悉一下
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    #include <queue>
    #include <vector>
    #include <iomanip>
    #include <math.h>
    #include <map>
    using namespace std;
    #define FIN     freopen("input.txt","r",stdin);
    #define FOUT    freopen("output.txt","w",stdout);
    #define INF     0x3f3f3f3f
    #define lson    l,m,rt<<1
    #define rson    m+1,r,rt<<1|1
    typedef long long LL;
    
    const int MAXN=1005;
    int dp[MAXN][MAXN];
    int value[MAXN];
    int volume[MAXN];
    
    int main()
    {
        //FIN
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int n,v;
            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=0;j<=v;j++)
                {
                    if(j>=volume[i])
                        dp[i][j]=max(dp[i-1][j],dp[i-1][j-volume[i]]+value[i]);
                    else
                        dp[i][j]=dp[i-1][j];
                }
            printf("%d
    ",dp[n][v]);
        }
    
    }
    

      





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  • 原文地址:https://www.cnblogs.com/Hyouka/p/5730984.html
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