Poj1611The Suspects

A - The Suspects

Time Limit: 1000 MS Memory Limit: 20000 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

Description

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others. 
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP). 
Once a member in a group is a suspect, all members in the group are suspects. 
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space. 
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output

For each case, output the number of suspects in one line.

Sample Input

100 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

并查集模板题，不过用的是没优化的

//Accepted	641 ms	272 KB	C++	1211 B	
#include <iostream>
#include <cstdio>

using namespace std;
const int maxn = 30005;
int father[maxn];

void init(int n)
{
	for(int i = 0; i < n; ++i)
		father[i] = i;
}
///查找一个节点所在的根节点
int serch(int v)
{
	if(father[v] == v)	return v;
    ///如果father[v] == v,v就是根，返回ｖ
	return serch(father[v]);　　　///路径压缩快　
	///否则继续查找根节点，此处是递归
}
///合并集合
void join(int x, int y)
{
    int fx = serch(x), fy = serch(y);
    if(fx != fy)
        father[fx] = fy;
}

int is_same(int x, int y)
{
    return (serch(x) == serch(y));
}

int main()
{
    int n, m;
    while(scanf("%d %d", &n, &m) != EOF && (n || m))
    {
        init(n);
        int t;
        for(int i = 0; i < m; ++i)
        {
            scanf("%d", &t);
            int a, b;
            scanf("%d", &a);
            t--;
            while(t--)
            {
                scanf("%d", &b);
                join(a, b);
            }
        }

        int res = 0;
        for(int i = 0; i < n; ++i)
        {
            if(serch(i) == serch(0))
                res++;
        }
        printf("%d
", res);
    }
    return 0;
}

　　

路径压缩可以快，而且只需改一行代码，很爽

//////16 ms	384 KB	C++	1221 B
int serch(int v)
{
	if(father[v] == v)	return v;
	return father[v] = serch(father[v]);　　　///路径压缩快　
}