题目大意:每个瓶子有一定的容积,以及一定的水量,问最少几个瓶子装满所有水,在此基础上还要最小化移动水的体积

第一问用贪心直接求
第二问转化成背包问题
设dp[i][j]表示前i桶水总容积为j的最多水量,这些水是不需要转移的
用总体积减去最多水量即为答案

代码:

 #include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
bool cmp(int x,int y) {return x>y;}
int n,lft,num,tot,ans;
int a[],b[],d[],dp[][];
int main()
{
cin>>n;
for(int i=;i<=n;i++) cin>>a[i],lft+=a[i];
for(int i=;i<=n;i++) cin>>b[i],d[i]=b[i];
sort(d+,d++n,cmp);
memset(dp,,sizeof(dp));
dp[][]=;
for(int i=;i<=n;i++)
{
tot+=d[i];
if(tot>=lft)
{
num=i;
break;
}
}
for(int i=;i<=n;i++)
for(int j=tot;j>=b[i];j--)
for(int k=;k<=num;k++)
dp[k][j]=max(dp[k][j],dp[k-][j-b[i]]+a[i]);
for(int i=lft;i<=tot;i++) ans=max(ans,dp[num][i]);
cout<<num<<" "<<lft-ans<<endl;
return ;
}

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