codeforces 417D. Cunning Gena 状压dp

题目链接

D. Cunning Gena

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A boy named Gena really wants to get to the "Russian Code Cup" finals, or at least get a t-shirt. But the offered problems are too complex, so he made an arrangement with his n friends that they will solve the problems for him.

The participants are offered m problems on the contest. For each friend, Gena knows what problems he can solve. But Gena's friends won't agree to help Gena for nothing: the i-th friend asks Gena xi rubles for his help in solving all the problems he can. Also, the friend agreed to write a code for Gena only if Gena's computer is connected to at least ki monitors, each monitor costs b rubles.

Gena is careful with money, so he wants to spend as little money as possible to solve all the problems. Help Gena, tell him how to spend the smallest possible amount of money. Initially, there's no monitors connected to Gena's computer.

Input

The first line contains three integers n, m and b (1 ≤ n ≤ 100; 1 ≤ m ≤ 20; 1 ≤ b ≤ 109) — the number of Gena's friends, the number of problems and the cost of a single monitor.

The following 2n lines describe the friends. Lines number 2i and (2i + 1) contain the information about the i-th friend. The 2i-th line contains three integers xi, ki and mi (1 ≤ xi ≤ 109; 1 ≤ ki ≤ 109; 1 ≤ mi ≤ m) — the desired amount of money, monitors and the number of problems the friend can solve. The (2i + 1)-th line contains mi distinct positive integers — the numbers of problems that the i-th friend can solve. The problems are numbered from 1 to m.

Output

Print the minimum amount of money Gena needs to spend to solve all the problems. Or print -1, if this cannot be achieved.

Sample test(s)

input

2 2 1
100 1 1
2
100 2 1
1

output

202

input

3 2 5
100 1 1
1
100 1 1
2
200 1 2
1 2

output

205

input

1 2 1
1 1 1
1

output

-1

直接状态压缩就可以， 具体看代码。 inf要超级大才可以过。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e18+;
const ll inf = 5e18+;
const int dir[][] = { {-, }, {, }, {, -}, {, } };
struct node
{
    int x, k, c;
    bool operator < (node a) const
    {
        return k<a.k;
    }
}a[];
ll dp[(<<)];
int main()
{
    int n, m, b, t, x;
    cin>>n>>m>>b;
    int ed = (<<m)-;
    for(int i = ; i<=n; i++) {
        cin>>a[i].x>>a[i].k>>t;
        while(t--) {
            scanf("%d", &x);
            a[i].c |= <<(x-);
        }
    }
    sort(a+, a++n);
    for(int i = ; i<=ed; i++) {
        dp[i] = inf;
    }
    ll ans = inf;
    for(int i = ; i<=n; i++) {
        for(int j = ; j<=ed; j++) {
            dp[j|a[i].c] = min(dp[j|a[i].c], dp[j]+a[i].x);
        }
        ans = min(ans, 1LL*a[i].k*b+dp[ed]);
    }
    if(ans == inf) {
        puts("-1");
    } else {
        cout<<ans<<endl;
    }
    return ;
}