Codeforces 417 D. Cunning Gena

最新推荐文章于 2018-07-19 11:59:48 发布

转载最新推荐文章于 2018-07-19 11:59:48 发布 · 88 阅读

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原文链接：http://www.cnblogs.com/blfbuaa/p/7323361.html

本文介绍了一道编程题的解决思路，通过状态压缩动态规划（DP）来找到解决所有比赛问题所需的最小花费。考虑了朋友帮助解决问题的成本及所需额外显示器的数量。

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按monitor排序，然后状压DP。。。

。

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 thei-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

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

long long int n,m,b;
long long int dp[ (1<<20)+10 ];

struct FD
{
    long long int x,k,pb;
}fd[120];

bool cmp(FD a,FD b)
{
    return a.k<b.k;
}

int main()
{
    scanf("%d%d%d",&n,&m,&b);
    for(int i=0;i<n;i++)
    {
        long long int xi,ki,mi,problem=0;
        cin>>xi>>ki>>mi;
        for(int j=0;j<mi;j++)
        {
            int a;
            scanf("%d",&a);
            a--;
            problem|=(1<<a);
        }
        fd[i]=(FD){xi,ki,problem};
    }
    sort(fd,fd+n,cmp);
    int state=(1<<m)-1;
    for(int i=0;i<=state;i++) dp[i]=(long long int )(1LL<<60);
    dp[0]=0;
    long long int ans=(long long int )(1LL<<60);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<=state;j++)
        {
            dp[j|fd[i].pb]=min(dp[j|fd[i].pb],dp[j]+fd[i].x);
        }
        ans=min(ans,dp[state]+fd[i].k*b);
    }
    if(ans==(long long int )(1LL<<60)) ans=-1;
    cout<<ans<<endl;
    return 0;
}

转载于:https://www.cnblogs.com/blfbuaa/p/7323361.html