ACM-ICPC 2018 南京赛区网络预赛 E-AC Challenge

Dlsj is competing in a contest with 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) si problems, the p1,p2,...pi,si-th problem before (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 i-th problem on t-th minute(or the t-th problem he solve is problem i), he can get t × ai + bi points. (∣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 n, which is the number of problems.

Then follows nn lines, the i-th line contains si + 3 integers ai,bi,si,p1,p2,...,pi 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 × 5 + 6 = 11 points.

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

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

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

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

So he can get 11 + 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

样例输出1

样例输入2

1
-100 0 0

样例输出2

Hint

状态压缩记忆化dp

#include <iostream>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <map>
#include <algorithm>
#include <queue>
#include <set>
#include <cmath>
#include <sstream>
#include <stack>
#include <fstream>
#include <ctime>
#pragma warning(disable:4996);
#define mem(sx,sy) memset(sx,sy,sizeof(sx))
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-8;
const double PI = acos(-1.0);
const ll llINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
using namespace std;
//#define pa pair<int, int>
//const int maxn = 1000005;
//const int mod = 1e9 + 7;int n;
int a[25], b[25];
vector<int> pre[25];
int dp[22][1 << 20];
int vis[22];
bool judge(int x, int state) {if (!vis[x]) {int flag = 0;for (int i = 0; i < pre[x].size(); i++) {if ((state & (1 << pre[x][i])) == 0) {flag = 1; break;}}if (flag)return 0;return 1;}return 0;
}
int dfs(int t, int now) {if (dp[t][now] != -1) return dp[t][now];int ret = 0;for (int i = 1; i <= n; i++) {int curget = (t + 1) * a[i] + b[i];if (judge(i, now)) {vis[i] = 1;ret = max(ret, dfs(t + 1, now | (1 << i)) + curget);vis[i] = 0;}}dp[t][now] = ret;return ret;
}
int main() {while (~scanf("%d", &n)) {mem(dp, -1);mem(vis, 0);int s;for (int i = 1, s; i <= n; i++) {pre[i].clear();scanf("%d%d%d", &a[i], &b[i], &s);for (int j = 1, temp; j <= s; j++) {scanf("%d", &temp);pre[i].push_back(temp);}}int ans = dfs(0, 0);printf("%d\n", ans);}
}