hdu 2660 Accepted Necklace (二维背包 简单题)
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2660
解题报告:题目中对每个物品的取舍包含两个方面的因素 1、自己选择的宝石种类 2、妈妈所能接受的最大的重量。所以是二维背包的类型的问题
由于每个物品只能使用一次,求最大的价值,所以转换为01背包。即:for(int j=use;j>=0;j++)
code:
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[1005][1005],val[1005],w[1005];int t,total,use,weight;int main(){scanf("%d",&t);while(t--){memset(dp,0,sizeof(dp));scanf("%d%d",&total,&use);for(int i=1;i<=total;i++)scanf("%d%d",&val[i],&w[i]);scanf("%d",&weight);for(int i=1;i<=total;i++)for(int j=weight;j>=w[i];j--)for(int k=use;k>=1;k--)dp[j][k]=max(dp[j][k],dp[j-w[i]][k-1]+val[i]);printf("%d\n",dp[weight][use]);}return 0;}
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