ACM-ICPC 2018 南京赛区网络预赛 E AC Challenge（状压dp）

Dlsj is competing in a contest with n (0 < n \le 20)n(0<n≤20) problems. And he knows the answer of all of these problems.

However, he can submit ii-th problem if and only if he has submitted (and passed, of course) s_isi​ problems, the p_{i, 1}pi,1​-th, p_{i, 2}pi,2​-th, ......, p_{i, s_i}pi,si​​-th problem before.(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 ii-th problem on tt-th minute(or the tt-th problem he solve is problem ii), he can get t \times a_i + b_it×ai​+bi​ points. (|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, nn, which is the number of problems.

Then follows nn lines, the ii-th line contains s_i + 3si​+3 integers, a_i,b_i,s_i,p_1,p_2,...,p_{s_i}ai​,bi​,si​,p1​,p2​,...,psi​​as 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 \times 5 + 6 = 111×5+6=11 points.

On the second minute, Dlsj submitted the second problem, and get 2 \times 4 + 5 = 132×4+5=13points.

On the third minute, Dlsj submitted the third problem, and get 3 \times 3 + 4 = 133×3+4=13 points.

On the forth minute, Dlsj submitted the forth problem, and get 4 \times 2 + 3 = 114×2+3=11 points.

On the fifth minute, Dlsj submitted the fifth problem, and get 5 \times 1 + 2 = 75×1+2=7 points.

So he can get 11+13+13+11+7=5511+13+13+11+7=55points 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

ACM-ICPC 2018 南京赛区网络预赛

题意：给你n(n<=20)门课，每门课需要1单位时间学习。每门课有k门前置课程。（学完所有前置课程才能学这门课）你在第t单位学习了第i门课，将会获得收益a[i]*t+b[i](|a|,|b|<=1e9)。你可以最多学习n门课，求能获得的最大收益（全都不学收益为0）。

思路：一眼过去就是状压dp啊。。。这两天刚复习完状压dp，没有犹豫，拿到这题迅速就1A了。

n比较小，直接二进制表示哪门课做了没做。

对于状态i，若(i&(1<<j)) ==1 就可以枚举当前要学习这第j+1门课了。tmp=i^(1<<j)表示没学这门课的状态（不合法跳过）。

然后看tmp|c[j]是否==tmp即可。（c[j]为前置课程的二进制形式），然后dp取最大值就行了。

代码：

#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<string>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define ll long long
using namespace std;
const int mo=1e9+7;
const int maxn=1<<20;
const ll inf=0x3f3f3f3f3f3f3f3fLL;
int n,m,k,T;
ll dp[maxn],ans,tmp,sum;
ll a[25],c[25],b[25];
ll fcount(int x)
{
 ll s=0;
 while(x){
  s++;
  x&=(x-1);
 }
    return s;
}
int main() {
     int T;
     while(scanf("%d",&n)!=EOF)
     {
         ans=0;sum=0;
         for(int i=0;i<n;i++)
         {
             scanf("%lld%lld%d",&a[i],&b[i],&k);
             c[i]=0;
             for(int j=0;j<k;j++)
             {
                 int x;
                 scanf("%d",&x);
                 x--;
                 c[i]|=(1<<x);
             }
         }
         m=1<<n;
         for(int i=0;i<m;i++)
         dp[i]=-inf;//数据水了，这里改成-1也能过，实际上-1不一定不合法
         dp[0]=0;
         for(int i=1;i<m;i++)
         {
             for(int j=0;j<n;j++)
             if(i&(1<<j)){
                 int tmp=i^(1<<j);
                 if(dp[tmp]==-inf)continue;
                 if((tmp|c[j])==tmp)
                 {
                     ll t=fcount(i);
                     ans=max(ans,dp[i]);
                     dp[i]=max(dp[i],dp[tmp]+a[j]*t+b[j]);
                     ans=max(ans,dp[i]);
                 }
             }
         }
         printf("%lld\n",ans);
     }
    return 0;
}