ACM-ICPC2018南京网络赛 AC Challenge（一维状压dp）

AC Challenge

•  30.04%
•  1000ms
•  128536K

 

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 imes 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 imes 5 + 6 = 111×5+6=11 points.

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

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

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

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

So he can get 11+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

ACM-ICPC 2018 南京赛区网络预赛

 

 

 

状压dp。

 

#include <bits/stdc++.h>
#define MAX 21
typedef long long ll;
using namespace std;
const int INF = 0x3f3f3f3f;

ll a[MAX],b[MAX];
ll dp[1<<20];
vector<int> v[MAX];

int main(void)
{
    int n,num,temp,i,j,k;
    scanf("%d",&n);
    for(i=1;i<=n;i++) {
        scanf("%lld %lld",&a[i],&b[i]);
        scanf("%d",&num);
        while(num--) {
            scanf("%d",&temp);
            v[i].push_back(temp);
        }
    }
    memset(dp,0,sizeof(dp));
    for(i=0;i<(1<<n);i++){
        int f=0;
        for(j=1;j<=n;j++){
            if(!((1<<(j-1))&i)) continue;
            for(k=0;k<v[j].size();k++){
                if(!((1<<(v[j][k]-1))&i)){
                    f=1;
                    break;
                }
            }
            if(f==1) break;
        }
        if(f==1) continue;
        for(j=1;j<=n;j++){
            if(!((1<<(j-1))&i)) continue;
            int S=i;
            int c=0;
            while(S){
                if(S&1) c++;
                S>>=1;
            }
            dp[i]=max(dp[i],dp[i^(1<<(j-1))]+c*a[j]+b[j]);
            //printf("(%d %d %d %lld)",i,c,j,dp[i]);
        }
    }
    printf("%lld
",dp[(1<<n)-1]);
    return 0;
}