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,...,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 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
因为n小于20 所以可以往状压上去想
每个题做与不做就是一种状态
枚举每种状态i 再枚举这个状态下最后做的题目j
通过状态转移方程 dp[i] = max(dp[i], dp[i ^ (1 << (j - 1))] + t * a[j] + b[j]
其中t是i中1的个数，也就是第j题如果最后做对应的时间
对于每一种状态i 要判断他这种状态所表示的所有做的题目成立不成立 也就是他这种状态是否成立
也就是说他所有是1的题目 都要判断是否满足他的要求
要求也可以转换为状态来存
另外： maxn = 25 会MLE， maxn = 21就过了
过程中忘记对i这个状态是否成立进行判断了
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#include<set>
//#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int maxn = 21;
struct pro{
int a, b, s;
int must;
}problems[maxn];
int dp[1 << maxn], n;
void init()
{
for(int i = 0; i < maxn;i++){
problems[i].must = 0;
}
memset(dp, 0, sizeof(dp));
}
int main()
{
while(scanf("%d", &n) != EOF){
init();
for(int i = 1; i <= n; i++){
scanf("%d%d%d", &problems[i].a, &problems[i].b, &problems[i].s);
for(int j = 0; j < problems[i].s; j++){
int p;
scanf("%d", &p);
problems[i].must |= (1 << (p - 1));
}
}
int ans = 0;
for(int i = 0; i < (1 << n); i++){
bool flag = true;
for(int j = 1; j <= n; j++){
if((i & (1 << (j - 1))) == 0){
continue;
}
if((i & problems[j].must) != problems[j].must){
flag = false;
break;
}
}
if(!flag) continue;
for(int j = 1; j <= n; j++){
if((i & (1 << (j - 1))) == 0){
continue;
}
int t = 0, tmp = i;
while(tmp){
if(tmp & 1)
t++;
tmp >>= 1;
}
dp[i] = max(dp[i], dp[i ^ (1 << (j - 1))] + t * problems[j].a + problems[j].b);
//ans = max(ans, dp[i]);
}
}
printf("%d
", dp[(1<<n) - 1]);
}
return 0;
}