URAL 1326. Bottle Taps(简单的状压dp)

题目不太好读懂，就是先给你一个n代表要从n个物品中买东西，然后告诉你这n个东西的单位价格，在给你m个集合的情况。就是每一个结合中有x件物品。他们合起来买的价格是k。这x件物品依次是：p1……px。之后给你一个kk，表示你要买的物品的编号。

让你求出来怎样花费最少的钱买到要求的序列。

20。能够状压啊。注意一開始的时候先把单位价格的状态处理出来。。。之后就是水题了啊。

1326. Bottle Taps

Time limit: 3.0 second

Memory limit: 64 MB

Programmer Petrov has a hobby to collect beer-bottle taps. There’s nothing unusual — he knows hundreds of programmers that like beer. And they collect taps, too. Not everyone, but some of them.

Frankly speaking, he has bought a part of his collection. But unfortunately he hasn’t got some rare taps to complete his collection. He has found some programmers over the Internet that are ready to
sell him these taps. Some of the programmers sell the taps in sets with big discounts.

It’s left to find an optimal offer. Petrov can explain to his wife why he is to store the taps but he won’t be able to prove why he is to spend money for the collection. So he is to buy the taps as
cheap as possible.

Petrov has written down all the variants and has started thinking. There’s no way to find out the solution of the problem without a program!

Input

The first line contains an integer N, an amount of available taps (1 ≤ N ≤ 20). The following Nlines contain prices of bottles with the taps if one buys them
in stores. The next line contains an integer M (0 ≤ M ≤ 100) — an amount of offers to sell the taps. The following M lines describe the sets. The first number of each line is the price of the set and the second one is
the amount of taps in the set. Then there are numbers of the taps in the set (each number lies in the range from 1 to N). The numbers in a set are unique. All the prices are positive integers and do not exceed 1000. The last line begins with the amount
of taps that Petrov plans to buy. Then their numbers follow separated by spaces. These numbers are unique, too.

Output

Output the minimal sum of money that Petrov should spend on obtaining the necessary taps.

Sample

input	output
4 10 11 12 13 3 17 2 1 3 25 3 2 3 4 15 2 3 4 3 1 3 4	25

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL __int64
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898

const int maxn = 1010;

using namespace std;

int dp[1<<22];
int num[maxn];
int f[maxn];
int st[maxn];
int n;

void dfs(int x, int y, int z)
{
    if(x == n)
    {
        dp[y] = z;
        return;
    }
    dfs(x+1, y, z);
    dfs(x+1, y|(1<<x), z+num[x]);
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i = 0; i < (1<<n); i++) dp[i] = INF;
        for(int i = 0; i < n; i++) scanf("%d",&num[i]);
        dfs(0, 0, 0);
        int m;
        scanf("%d",&m);
        for(int i  = 0; i < m; i++)
        {
            scanf("%d",&f[i]);
            int x;
            scanf("%d",&x);
            int sum = 0;
            int tmx;
            for(int j = 0; j < x; j++)
            {
                scanf("%d",&tmx);
                sum |= (1<<(tmx-1));
            }
            st[i] = sum;
        }
        int kk;
        scanf("%d",&kk);
        int ans = 0;
        int p;
        for(int i = 0; i < kk; i++)
        {
            scanf("%d", &p);
            ans |= (1<<(p-1));
        }
        int Min = INF;
        for(int i = 0; i < (1<<n); i++)
        {
            if((i&ans) == ans) Min = min(Min, dp[i]);
            for(int j = 0; j < m; j++) dp[i|st[j]] = min(dp[i|st[j]], dp[i]+f[j]);
        }
        printf("%d\n",Min);
    }
    return 0;
}