POJ2443
Set Operation
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 2679 | Accepted: 1050 |
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
Hint
The input may be large, and the I/O functions (cin/cout) of C++ language may be a little too slow for this problem.
Source
POJ Monthly,Minkerui
主要是简单的状态压缩，由于数字最大就是10000,而int最大是32位不到，所以切割成10000/30个，分成40份好了。
之后进行状态压缩，奇妙的二进制啊。
经过这样的处理后，插入是常数，而推断一个数是否在某一个集合也是常数时间。所以每个查询推断他是否在集合中也就是O(n)。
典型的空间换时间。
一開始我开了一个1000*10000的数组。memset斗能够导致TLE了。
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct st{
int u[400];
void add(int n){
int i=n/30,j=n%30;
u[i]|=(1<<j);
}
bool in(int n){
int i=n/30,j=n%30;
return u[i]&(1<<j);
}
void del(int n){
int i=n/30,j=n%30;
u[i]&=~(1<<j);
}
void init(){
memset(u,0,sizeof(u));
}
};
st v[1001];
int main()
{
int n,m,i,s,t;
while(scanf("%d",&n)!=EOF){
for(i=0;i<n;i++){
v[i].init();
scanf("%d",&m);
while(m--){
scanf("%d",&s);
v[i].add(s);
}
}
scanf("%d",&m);
while(m--){
scanf("%d%d",&s,&t);
for(i=0;i<n;i++){
if(v[i].in(s)&&v[i].in(t)){
break;
}
}
if(i==n){
puts("No");
}else{
puts("Yes");
}
}
}
return 0;
}