• HDU 2602 Bone Collector (01背包问题)


    原题代号:HDU 2602
    原题描述:
    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
     
    题目大意:骨头收集者在收集骨头,有容量为v的袋子,n个骨头,给出每个骨头的价值和体积。问最大能收集的价值是多少。(01背包问题)
     
    解法一:时间,空间复杂度均为O(n2)
    # include <stdio.h>
    # include <string.h>
    # include <stdlib.h>
    # include <iostream>
    # include <fstream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <math.h>
    # include <algorithm>
    using namespace std;
    # define pi acos(-1.0)
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define For(i,n,a) for(int i=n; i>=a; --i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define Fo(i,n,a) for(int i=n; i>a ;--i)
    typedef long long LL;
    typedef unsigned long long ULL;
    
    int a[1000+5],b[1000+5],dp[1000+5][1000+5];
    
    int main()
    {
        int t,n,v;
        cin>>t;
        while(t--)
        {
            cin>>n>>v;
            FOR(i,1,n)cin>>b[i];
            FOR(i,1,n)cin>>a[i];
            mem(dp,0);
            FOR(i,1,n)FOR(j,0,v)
            {
                if(a[i]<=j)dp[i][j]=max(dp[i-1][j],dp[i-1][j-a[i]]+b[i]);
                else dp[i][j]=dp[i-1][j];
            }
            cout<<dp[n][v]<<endl;
        }
        return 0;
    }

    解法二:时间复杂度O(n2),空间复杂度O(n)

    # include <stdio.h>
    # include <string.h>
    # include <stdlib.h>
    # include <iostream>
    # include <fstream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <math.h>
    # include <algorithm>
    using namespace std;
    # define pi acos(-1.0)
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define For(i,n,a) for(int i=n; i>=a; --i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define Fo(i,n,a) for(int i=n; i>a ;--i)
    typedef long long LL;
    typedef unsigned long long ULL;
    
    int a[1000+5],b[1000+5],dp[1000+5];
    
    int main()
    {
        int t,n,v;
        cin>>t;
        while(t--)
        {
            int ans=0;
            cin>>n>>v;
            FOR(i,1,n)cin>>b[i];
            FOR(i,1,n)cin>>a[i];
            mem(dp,0);
            FOR(i,1,n)For(j,v,0)
            {
                if(a[i]<=j)dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
            }
            cout<<dp[v]<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/teble/p/7191330.html
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