Dlsj is competing in a contest with n(0<n≤20)n (0 < n \le 20)n(0<n≤20) problems. And he knows the answer of all of these problems.
However, he can submit iii-th problem if and only if he has submitted (and passed, of course) sis_isi problems, the pi,1p_{i, 1}pi,1-th, pi,2p_{i, 2}pi,2-th, ........., pi,sip_{i, s_i}pi,si-th problem before.(0<pi,j≤n,0<j≤si,0<i≤n)(0 < p_{i, j} \le n,0 < j \le s_i,0 < i \le n)(0<pi,j≤n,0<j≤si,0<i≤n) After the submit of a problem, he has to wait for one minute, or cooling down time to submit another problem. As soon as the cooling down phase ended, he will submit his solution (and get "Accepted" of course) for the next problem he selected to solve or he will say that the contest is too easy and leave the arena.
"I wonder if I can leave the contest arena when the problems are too easy for me."
"No problem."
—— CCF NOI Problem set
If he submits and passes the iii-th problem on ttt-th minute(or the ttt-th problem he solve is problem iii), he can get t×ai+bit \times a_i + b_it×ai+bi points. (∣ai∣,∣bi∣≤109)(|a_i|, |b_i| \le 10^9)(∣ai∣,∣bi∣≤109).
Your task is to calculate the maximum number of points he can get in the contest.
Input
The first line of input contains an integer, nnn, which is the number of problems.
Then follows nnn lines, the iii-th line contains si+3s_i + 3si+3 integers, ai,bi,si,p1,p2,...,psia_i,b_i,s_i,p_1,p_2,...,p_{s_i}ai,bi,si,p1,p2,...,psias described in the description above.
Output
Output one line with one integer, the maximum number of points he can get in the contest.
Hint
In the first sample.
On the first minute, Dlsj submitted the first problem, and get 1×5+6=111 \times 5 + 6 = 111×5+6=11 points.
On the second minute, Dlsj submitted the second problem, and get 2×4+5=132 \times 4 + 5 = 132×4+5=13 points.
On the third minute, Dlsj submitted the third problem, and get 3×3+4=133 \times 3 + 4 = 133×3+4=13 points.
On the forth minute, Dlsj submitted the forth problem, and get 4×2+3=114 \times 2 + 3 = 114×2+3=11 points.
On the fifth minute, Dlsj submitted the fifth problem, and get 5×1+2=75 \times 1 + 2 = 75×1+2=7 points.
So he can get 11+13+13+11+7=5511+13+13+11+7=5511+13+13+11+7=55 points in total.
In the second sample, you should note that he doesn't have to solve all the problems.
样例输入1
5 5 6 0 4 5 1 1 3 4 1 2 2 3 1 3 1 2 1 4
样例输出1
55
样例输入2
1 -100 0 0
样例输出2
0
题意：
给你n个题目，每个题目有三个权值，ai,bi,si，之后给你si个数字p1,p2...psi，如果要做出这道题，必须先把这si个题目先做完，每道题目需要1分钟去做，如果你在第t分钟做出了这道题，那么你将会获得ai*ti+bi点分数，要你求你会获得的最大分数。
做法：
机房里的大神说是一眼题，毕竟n只有20，稍微有点经验的人一看就知道是状压dp，这题是队友做的，我虽然也很想负责dp但是无奈脑子不够用，还是多做题。。其实还挺喜欢dp的。
解释在下面的代码里，对状压dp有了解的话应该能很快明白的。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
typedef long long ll;
const int maxn=(1<<20);
const int inf=0x3f3f3f3f;
ll read(){
ll ans=0; char last=' ',ch=getchar();
while(ch<'0' || ch>'9')last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10ll+ch-'0',ch=getchar();
if(last=='-')ans=-ans; return ans;
}
ll dp[maxn+5];
ll a[25],b[25],s[25];
struct node{
ll aim,ti;
node(ll aim,ll ti):aim(aim),ti(ti){}
};
int ck(ll aim,ll tmp,int i){
if(aim&tmp) return 1;//如果已经做过了这道题 不用再做
int x=aim&s[i];
if(x!=s[i]) return 1;//如果没达到要做这道题的条件（即前提的s[i]没做完）
return 0;
}
int main(){
int n=read(),x,y;
for(int i=0;i<=maxn;i++)
dp[i]=-inf;
for(int i=1;i<=n;i++){
a[i]=read(); b[i]=read(); x=read();
ll now=0;
while(x--){
ll y=read();
now+=(1<<(y-1));
}
s[i]=now;//预处理要做这道题的条件
}
queue<node> q;
q.push(node(0,0));
dp[0]=0;
while(!q.empty()){
node _=q.front(); q.pop();
ll aim=_.aim,ti=_.ti+1;
for(int i=1;i<=n;i++){
int tmp= (1<<(i-1)) ;
if(ck(aim,tmp,i)) continue;
int y=aim+tmp;
if(dp[y]<dp[aim]+ti*a[i]+b[i]){//如果这是更优的解
dp[y]=dp[aim]+ti*a[i]+b[i];
q.push(node(y,ti));
}
}
}
ll ans=0;
for(int i=0;i<=maxn;i++){
ans=max(ans,dp[i]);
}
printf("%lld\n",ans);
}