Junk-Mail Filter
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1785 Accepted Submission(s): 521
Problem Description
Recognizing junk mails is a tough task. The method used here consists of two steps:
1) Extract the common characteristics from the incoming email.
2) Use a filter matching the set of common characteristics extracted to determine whether the email is a spam.
We want to extract the set of common characteristics from the N sample junk emails available at the moment, and thus having a handy data-analyzing tool would be helpful. The tool should support the following kinds of operations:
a) “M X Y”, meaning that we think that the characteristics of spam X and Y are the same. Note that the relationship defined here is transitive, so
relationships (other than the one between X and Y) need to be created if they are not present at the moment.
b) “S X”, meaning that we think spam X had been misidentified. Your tool should remove all relationships that spam X has when this command is received; after that, spam X will become an isolated node in the relationship graph.
Initially no relationships exist between any pair of the junk emails, so the number of distinct characteristics at that time is N.
Please help us keep track of any necessary information to solve our problem.
Input
There are multiple test cases in the input file.
Each test case starts with two integers, N and M (1 ≤ N ≤ 105 , 1 ≤ M ≤ 106), the number of email samples and the number of operations. M lines follow, each line is one of the two formats described above.
Two successive test cases are separated by a blank line. A case with N = 0 and M = 0 indicates the end of the input file, and should not be processed by your program.
Output
For each test case, please print a single integer, the number of distinct common characteristics, to the console. Follow the format as indicated in the sample below.
Sample Input
5 6
M 0 1
M 1 2
M 1 3
S 1
M 1 2
S 3
3 1
M 1 2
0 0
Sample Output
题目分析:
题目的意思大概就是 有N 封邮件, 编号 0 -> N-1, 然后有2种操作, M : 合并操作, 将 2 种邮件合并为一种.
S : 分离操作, 将一封邮件独立出去, 单独占一个集合.
最后题目要求统计 集合的 个数. 从这里可以很容易的看出, 这是一个 并查集的题目, 不过按朴素方法来做的话, 一般都会 TLE.
加上数据量很大 , 不要使用 cin , 会超时. 而且一般来说 G ++ 和 C++ 在处理大量数据的时候会有1倍的时差 !!! 所以一般建议使用
C++ 提交代码.
先看代码 :
/*
MiYu原创, 转帖请注明 : 转载自 ______________白白の屋
http://www.cnblog.com/MiYu
Author By : MiYu
Test : 1
Program : 2473
*/
#include <iostream>
#include <algorithm>
using namespace std;
int set[1350005];
int a[125000];
int N,M;
int inline find ( int x )
{
return x != set[x] ? set[x] = find ( set[x] ) : set[x];
}
void inline merge ( int x, int y )
{
x = find ( x );
y = find ( y );
if ( x == y ) return ;
set[x] = y;
}
int main ()
{
int ca = 1;
while ( scanf ( "%d%d",&N,&M ), M || N )
{
for ( int i = 0; i < N; ++ i )
set[i] = i+N;
for ( int i = N; i <= N + N + M; ++ i )
set[i] = i; // <------------------------------这是关键, 虽然空间的消耗比较大, 但是节省了大量时间, 这样处理的目的是将0 -> N-1 的 节点处理成叶子节点
// 这样在对这些 节点做 S 操作的时候就不会影响到其他的节点, 而 find 操作是带路径压缩的, 所以就保证了我们所 // 有要处理的节点一直是叶子节点 !!!
int sep = N + N;
int x,y; char ch[5];
for ( int i = 0; i != M; ++ i )
{
scanf ( "%s",ch );
switch ( ch[0] )
{
case 'M': scanf ( "%d%d",&x,&y ); merge ( x,y ); break;
case 'S': scanf ( "%d",&x ); set[x] = sep ++; break; //初始化操作就是为这一步准备的
}
}
for ( int i = 0; i != N; ++ i )
a[i] = find ( i );
sort ( a, a + N );
int nCount = 1;
for ( int i = 1; i < N; ++ i )
if ( a[i] != a[i-1] ) nCount ++;
printf("Case #%d: %d\n",ca++,nCount);
}
return 0;
}