ACM-ICPC 2018 南京賽區網絡預賽 E題

ACM-ICPC 2018 南京賽區網絡預賽 E題

題目連接： https://nanti.jisuanke.com/t/30994

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,...,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 \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.

樣例輸入1複製

5
5 6 0
4 5 1 1
3 4 1 2
2 3 1 3
1 2 1 4

樣例輸出1複製

55

樣例輸入2複製

1
-100 0 0

樣例輸出2複製

0

題目來源

ACM-ICPC 2018 南京賽區網絡預賽

 

看到了n<20, 很顯然就是要狀態壓縮了。

而後那個前提條件，其實就是狀態轉移時候的條件了，只要判斷一下就能夠了。

 

狀態壓縮就是用一個二進制數來表示當前的狀態，這個數須要有n個二進制位，若是爲0表示沒有submit這個題，1表示submit了。

dp[i]表示到狀態i得到的最多points. 根據i有多少1就知道如今的時間了（每分鐘submit一個）。

 

具體看代碼吧。

​

 

#include <bits/stdc++.h>
using namespace std;
​
​
int bit[22];
​
int a[22];
int b[22];
int state[22];
​
long long dp[1<<20];
int numbit[1<<20];
const long long INF = 1000000000000000LL;
​
int main() {
  bit[0] = 1;
  for (int i = 1; i < 22; i++)
    bit[i] = bit[i-1]<<1;
  numbit[0] = 0;
  for (int i = 1; i < bit[20]; i++) {
    numbit[i] = 1 + numbit[i&(i-1)];
  }
  int n;
  while(scanf("%d", &n) == 1) {
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &a[i], &b[i]);
      int s;
      scanf("%d", &s);
      int tmp = 0;
      state[i] = 0;
      while (s--) {
        scanf("%d", &tmp);
        state[i] |= bit[tmp-1];
      }
    }
      dp[0] = 0;
      for (int i = 1; i < bit[n]; i++)dp[i] = -INF;
      long long ans = 0;
      for (int i = 0; i < bit[n]; i++) {
        if (dp[i] == -INF)continue;
        ans = max(ans, dp[i]);
        for (int j = 0; j < n; j++) {
          if (i & bit[j])continue;
          if ((i&state[j]) != state[j])continue;
          dp[i|bit[j]] = max(dp[i|bit[j]], dp[i] + (long long)(numbit[i]+1)*a[j] + b[j]);
        }
      }
      cout<<ans<<endl;
​
  }
  return 0;
}

　　

 

 https://kuangbin.github.io/2018/09/01/2018-ACM-ICPC-Nanjing-online-E/