题意

大赛将至，摆在你面前的是n道题目，第 i(1 ≤ i ≤ n) 道题目能提升 ai 点智力值，代码量为 bi KB，无聊值为 ci ，求至少提升m点智力值的情况下，所做题目代码量之和*无聊值之和最小为多少。

题解

像最小乘积树那么做就可以了
就是吧一个k**(我不会拼)的过程改成DP而已
别的就直接上模板就可以了

感觉这种求乘积的都可以这么做啊
很套路的感觉啊

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long LL;
const LL MAX=1LL<<55;
const LL N=405;
LL n,m;
LL a[N],b[N],c[N];
LL d[N];
struct node{LL x,y;}s[N];
LL ans=MAX;
LL f[N][1005];//前i个，达到标准的最小值 
bool g[N][1005];
node DP ()
{for (LL u=0;u<=n;u++)for (LL i=0;i<=800;i++)f[u][i]=MAX;f[0][0]=0;for (LL u=0;u<n;u++)for (LL i=0;i<=800;i++)if (f[u][i]!=MAX){if (f[u][i]+d[u+1]<f[u+1][i+a[u+1]]){f[u+1][i+a[u+1]]=f[u][i]+d[u+1];g[u+1][i+a[u+1]]=true;}if (f[u][i]<f[u+1][i]){f[u+1][i]=f[u][i];g[u+1][i]=false;}}LL k=-1;for (LL u=m;u<=800;u++)if (f[n][u]!=MAX&&(k==-1||f[n][u]<f[n][k])) k=u;
// if (k==-1) return ;node p;p.x=0;p.y=0;
/* printf("d:");for (int u=1;u<=n;u++) printf("%lld ",d[u]);printf("\n");printf("OZY:%lld\n",k);*/for (LL u=n;u>=1;u--)if (g[u][k]){k=k-a[u];p.x+=b[u];p.y+=c[u];}
// printf("YES:%lld %lld\n",p.x,p.y);ans=min(ans,p.x*p.y);return p;
}
LL mul (node x,node y,node z)
{LL x1=x.x-z.x,y1=x.y-z.y;LL x2=y.x-z.x,y2=y.y-z.y;return x1*y2-x2*y1;
}
void solve (node x,node y)
{LL yy=x.y-y.y,xx=y.x-x.x;for (LL u=1;u<=n;u++)    d[u]=b[u]*yy+c[u]*xx;node p=DP();if (mul(p,x,y)>=0) return ;solve(x,p);solve(p,y);
}
int main()
{scanf("%lld%lld",&n,&m);for (LL u=1;u<=n;u++)   scanf("%lld%lld%lld",&a[u],&b[u],&c[u]);for (LL u=1;u<=n;u++)   d[u]=b[u];node a=DP();for (LL u=1;u<=n;u++) d[u]=c[u];node b=DP();solve(a,b);printf("%lld\n",ans);return 0;
}