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复制
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复制
55
样例输入2复制
1 -100 0 0
样例输出2复制
0
题目来源
分析:由于n很小,所以可以枚举每道题目的是否做的状态。所以用状压dp来枚举每种状态。
例如:101可以有100和001这两种状态来得到。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node
{
ll a;
ll b;
ll len;
vector<ll> M;
}edge[30];
ll dp[1<<21];
int main()
{
ll n;
while(~scanf("%lld",&n))
{
memset(dp,0,sizeof(dp));
for(ll i=1;i<=n;i++)
{
scanf("%lld%lld%lld",&edge[i].a,&edge[i].b,&edge[i].len);
for(ll j=1;j<=edge[i].len;j++)
{
ll x;
scanf("%lld",&x);
edge[i].M.push_back(x);
}
}
for(ll k=0;k<=(1<<n)-1;k++)
{
ll flog=0;
for(ll i=1;i<=n;i++)
{
if(!((1<<(i-1))&k)) continue; ///当前的状态该位置对应的是0,所以可以不用考虑该位置。例如101,你只需要考虑第一题和第三题即可,不需要考虑第二题。
for(ll j=0;j<edge[i].len;j++)
{
if(!(k&(1<<(edge[i].M[j]-1))))
{
flog=1;
break;
}
}
}
if(flog)
continue;
for(ll i=1;i<=n;i++)
{
if(!((1<<(i-1))&k)) continue;
ll dex=k;
ll idex=0;
while(dex)
{
if(dex&1) idex++;
dex=dex>>1;
}
dp[k]=max(dp[k],dp[k-(1<<(i-1))]+edge[i].a*idex+edge[i].b);
// cout<<(k^(1<<(i-1)))<<"*"<<(k-(1<<(i-1)))<<endl;
}
}
// for(ll i=0;i<=(1<<n)-1;i++)
// cout<<i<<"->"<<dp[i]<<endl;
printf("%lld\n",dp[(1<<n)-1]);
}
return 0;
}
/*
5
5 6 0
4 5 1 1
3 4 1 2
2 3 1 3
1 2 1 4
*/