您的位置：首页 > 运维架构

poj 2443 Set Operation （位操作）

2013-09-27 21:13 162 查看
Set Operation

Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 2307 Accepted: 892
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
思路：

开一个a[32][10001]的数组就能保存所有信息了，并且能优化时间了。充分利用一个int能保存32个信息，把矩阵的每列都看做一个二进制数，但是这个数太大，所以需要分n/32+1个数来保存。每次操作时只要对相应的列进行&操作就够了。既节省了空间，有优化了时间，确实蛮经典的。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
//#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 10005
#define mod 1000000000
#define INF 0x3f3f3f3f
using namespace std;

int n,m,ans,q,flag;
int a[32][maxn];

int main()
{
int i,j,t,x,y,u,v;
while(~scanf("%d",&n))
{
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
{
scanf("%d",&m);
x=i/32;
t=i%32;
for(j=0;j<m;j++)
{
scanf("%d",&u);
y=u;
a[x][y]|=(1<<t);
}
}
m=n/32;
scanf("%d",&q);
for(i=1;i<=q;i++)
{
scanf("%d%d",&u,&v);
flag=0;
for(j=0;j<=m;j++)
{
if(a[j][u]&a[j][v])
{
flag=1;
break ;
}
}
if(flag) printf("Yes\n");
else printf("No\n");
}
}
return 0;
}
内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理 点击这里给我发消息
标签： 
相关文章推荐
新的分享
章节导航