POJ 2443 Set Operation (按位压缩)

Description

You are given N sets, the i-th set (represent by S(i)) have C(i) element (Here "set" isn't entirely the same as the "set" defined in mathematics, and a set may contain two same element). Every element in a set is represented by a positive number from 1 to 10000. Now there are some queries need to answer. A query is to determine whether two given elements i and j belong to at least one set at the same time. In another word, you should determine if there exist a number k (1 <= k <= N) such that element i belongs to S(k) and element j also belong to S(k).

Input

First line of input contains an integer N (1 <= N <= 1000), which represents the amount of sets. Then follow N lines. Each starts with a number C(i) (1 <= C(i) <= 10000), and then C(i) numbers, which are separated with a space, follow to give the element in the set (these C(i) numbers needn't be different from each other). The N + 2 line contains a number Q (1 <= Q <= 200000), representing the number of queries. Then follow Q lines. Each contains a pair of number i and j (1 <= i, j <= 10000, and i may equal to j), which describe the elements need to be answer.

Output

For each query, in a single line, if there exist such a number k, print "Yes"; otherwise print "No".

Sample Input

3
3 1 2 3
3 1 2 5
1 10
4
1 3
1 5
3 5
1 10

Sample Output

Yes
Yes
No
No

给你N(1<=N<=1000)个集合,每个集合里面有不超过10000个可以重复的元素
给你不超过10000次的询问,每个询问给你两个数,问有没有一个集合同时包含这两个数
思路:按照位进行压缩,因为对与一个数是否在集合中只有在或者不在两种情况于是用一个二进制位存就够了
这样大大节省了空间,一个int是32位,询问的数字最大是10000.我们按照30个一组的分块,一共大概要分400块
这很类似与硬盘的分块...然后我们用位运算维护更新下每一个集合,对于每个询问我们去枚举每一个集合就可以了

#include <iostream>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
const int BufferSize=1<<16;
char buffer[BufferSize],*head,*tail;
inline char Getchar() {
    if(head==tail) {
        int l=fread(buffer,1,BufferSize,stdin);
        tail=(head=buffer)+l;
    }
    return *head++;
}
inline int read() {
    int x=0,f=1;char c=Getchar();
    for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
    return x*f;
}
const int maxn = 1100;
const int Size = 30;
int n,m,q;
struct Set
{
    int st[400];
    void init (){for (int i=0;i<400;++i) st[i]=0;}
    int getBlk(int num){return num/Size;}
    int getIdx(int num){return num%Size;}
    void Add (int x){st[getBlk(x)]|=(1<<getIdx(x));}
    void Del (int x){st[getBlk(x)]&=~(1<<getIdx(x));}
    bool In (int x){return st[getBlk(x)]&(1<<getIdx(x));}
};
Set a[maxn];
int main()
{
    //freopen("de.txt","r",stdin);
    while (~scanf("%d",&n)){
        for (int i=0;i<maxn;++i)
            a[i].init();
        for (int i=0;i<n;++i){
            m=read();
            for (int j=0;j<m;++j){
                int x;
                x=read();
                a[i].Add(x);
            }
        }
        q=read();
        for (int i=0;i<q;++i){
            int x,y;
            x=read();y=read();
            bool f=false;
            for (int i=0;i<n;++i){
                if (a[i].In(x)&&a[i].In(y)){
                    f=true;break;
                }
            }
            if(f) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}