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  • hiho 第六周 01背包

    简单的01背包,没有报名,这周的没有权限提交

     1 #include<iostream>
     2 #include<memory.h>
     3 using namespace std;
     4 #define maxn 505
     5 #define maxm 100005
     6 int need[maxn],value[maxn];
     7 int dp[maxm];
     8 int main()
     9 {
    10     int n,m;
    11     while(cin>>n>>m)
    12     {
    13         memset(dp,0,sizeof(dp));
    14         for(int i = 1; i <= n; i++)
    15             cin>>need[i]>>value[i];
    16         for(int j = 0; j <= m; j++)
    17             dp[j] = 0;
    18         for(int i = 1; i <= n; i++)
    19             for(int j = m; j >= need[i]; j--)
    20                 dp[j] = max(dp[j],dp[j - need[i]] + value[i]);
    21         cout<<dp[m]<<endl;
    22     }
    23     return 0;
    24 }
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  • 原文地址:https://www.cnblogs.com/imLPT/p/3930568.html
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