【Codeforces 367D 】Sereja and Sets 位运算_codeforces sereja and sets-CSDN博客

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D. Sereja and Sets
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Sereja has m non-empty sets of integers A1, A2, …, Am. What a lucky coincidence! The given sets are a partition of the set of all integers from 1 to n. In other words, for any integer v (1 ≤ v ≤ n) there is exactly one set At such that . Also Sereja has integer d.

Sereja decided to choose some sets from the sets he has. Let’s suppose that i1, i2, …, ik (1 ≤ i1 < i2 < … < ik ≤ m) are indexes of the chosen sets. Then let’s define an array of integers b, sorted in ascending order, as a union of the chosen sets, that is, . We’ll represent the element with number j in this array (in ascending order) as bj. Sereja considers his choice of sets correct, if the following conditions are met:

b1 ≤ d; bi + 1 - bi ≤ d (1 ≤ i < |b|); n - d + 1 ≤ b|b|.
Sereja wants to know what is the minimum number of sets (k) that he can choose so that his choice will be correct. Help him with that.

Input
The first line contains integers n, m, d (1 ≤ d ≤ n ≤ 105, 1 ≤ m ≤ 20). The next m lines contain sets. The first number in the i-th line is si (1 ≤ si ≤ n). This number denotes the size of the i-th set. Then the line contains si distinct integers from 1 to n — set Ai.

It is guaranteed that the sets form partition of all integers from 1 to n.

Output
In a single line print the answer to the problem — the minimum value k at the right choice.

Sample test(s)
input
3 2 2
1 2
2 1 3
output
1
input
5 1 1
5 4 5 3 2 1
output
1
input
7 3 1
4 1 3 5 7
2 2 6
1 4
output
3
題意：；
题解：除去不合法的，then暴力count

#include<iostream>
#include<stdio.h>
using namespace std;
int n,m,d;
int num[25];
int no[1<<20];
int be[200000];
int main()
{
    scanf("%d%d%d",&n,&m,&d);
    for(int i=0;i<m;i++)
    {
        int aa;int bb;
        scanf("%d",&aa);
        for(int j=1;j<=aa;j++)
        {
            scanf("%d",&bb);
            be[bb]=i;
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(i>d) num[be[i-d]]--;
        num[be[i]]++;
        if(i>=d)
        {
            int sta=0;
            for(int j=0;j<m;j++)
            {   
                if(!num[j])
                sta|=(1<<j);
            }
            no[sta]=1;
        }
    }


    int ans=30;
    for(int j=(1<<m)-1;j>0;j--)
    {
        if(no[j])
        {
             for(int i=0;i<m;i++)
             {
                if(j&(1<<i))
                no[j^(1<<i)]=1;
             }
        }
        else ans=min(ans, __builtin_popcount(j));
    }

    printf("%d",ans);

}