- 1000ms
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AC Challenge
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
题意：给你n个作业，每完成一个作业的可以得到a*t+b的奖励，t表示完成的第几个作业，要完成第i个作业必须完成其他若干个作业，最后问你最后能得到的最大的奖励。
思路：毫不犹豫状压dp，dp[j]表示当前状态下可以得到的最大奖励，j中二进制中的01表示那个左右有没有被完成，
我的状态转移方程为
if((sta[i]&j)==sta[i]&&(j!=(j|(1<<(i-1)))))dp[j|(1<<(i-1))]=max(dp[j|(1<<(i-1))],dp[j]+get1(j|(1<<(i-1)))*a[i]+b[i]);
表示在状态j的时候，完成第i个作业需要的状态为sta[i],当状态j满足状态sta[i]且状态j下第i个作业还没有被完成下进行转移，
get1（）表示这个数的二进制中有多少个1，应为每个1表示那个作业完成了。
代码：
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<set>
#include<map>
#define ll long long
#define qq printf("QAQ\n");
using namespace std;
const int maxn=5+(1<<21);
const int inf=0x3f3f3f3f;
const ll linf=8e18+9e17;
const int mod=1e9+7;
const double e=exp(1.0);
const double pi=acos(-1);
const double eps=1e-6;
int sta[maxn];
ll dp[maxn]={0};
int get1(int x)
{
int ans=0;
while(x)
{
if(x&1)ans++;
x/=2;
}
return ans;
}
int main()
{
int t,n,last;
ll a[25],b[25];
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%lld%lld%d",&a[i],&b[i],&t);
sta[i]=0;
for(int j=1;j<=t;j++)
{
scanf("%d",&last);
sta[i]|=1<<(last-1);
}
}
memset(dp,-inf,sizeof dp);
dp[0]=0;
ll ans=0;
for(int j=0;j<=1<<n;j++)//枚举状态
for(int i=1;i<=n;i++)//枚举作业
{
if((sta[i]&j)==sta[i]&&(j!=(j|(1<<(i-1))))){
//状态满足 当前状态第i个作业没完成
dp[j|(1<<(i-1))]=max(dp[j|(1<<(i-1))],dp[j]+get1(j|(1<<(i-1)))*a[i]+b[i]);
}
}
int mx=(1<<n);
for(int i=0;i<=mx;i++)
ans=max(ans,(ll)dp[i]);
printf("%lld\n",ans);
}
return 0;
}