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  • 01背包

    <span style="color:#3333ff;">/*
    __________________________________________________________________________________________________
    *     copyright:   Grant Yuan                                                                     *
    *     algorithm:   01背包                                                                       *
    *     time     :   2014.7.18                                                                      *
    *_________________________________________________________________________________________________*
    F - 01背包
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
    Submit
     
    Status
    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 */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<functional> #include<queue> #include<stack> #include<cstdlib> using namespace std; int v[1001]; int p[1001]; int n,sv,t; int dp[2][1001]; int main() { cin>>t; while(t--){ cin>>n>>sv; for(int i=0;i<n;i++) cin>>p[i]; for(int i=0;i<n;i++) cin>>v[i]; memset(dp,0,sizeof(dp)); for(int i=0;i<n;i++) for(int j=0;j<=sv;j++) { if(j<v[i]) dp[(i+1)&1][j]=dp[i&1][j]; else dp[(i+1)&1][j]=max(dp[i&1][j],dp[i&1][j-v[i]]+p[i]); } cout<<dp[n&1][sv]<<endl; } return 0; } </span>


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  • 原文地址:https://www.cnblogs.com/bhlsheji/p/5347482.html
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