并查集——The Suspects

最新推荐文章于 2022-02-28 11:04:50 发布

原创最新推荐文章于 2022-02-28 11:04:50 发布 · 159 阅读

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本文介绍了一个基于并查集算法的SARS传播模拟问题，通过构建学生群体关系图，找出所有潜在的SARS嫌疑人。文章详细解释了算法实现过程，包括初始化并查集、合并关系和查找根节点等关键步骤。

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这道题也还是模板的套用加上在输入的地方做一些改动就可以
题意描述：
已知0是嫌疑犯的一个节点，和他有关系的（树建成后和它在一个树上的就是）也变成嫌疑人，题意就是求出嫌疑人的数量
解题思路：
题目中每一个样例都会有很多组关系，做法是让每组的第一个人依次和在同一数组中的后面节点合并（也就是建立关系（建树））要是不理解的话可以直接看代码代码，
题目：
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
Sample Output
4
1
1
AC代码：

#include<stdio.h>
int n,m,f[50050];
void init()
{
	int i;
	for(i=0; i<n; i++)
	  f[i]=i;
	  return ;
}
int getf(int v)
{
	if(f[v]==v)
	 return v;
	 else
	 {
	 	f[v]=getf(f[v]);
	 	return f[v];
	 }
}

void merge(int v,int u)
{
	int t1,t2;
	t1=getf(v);
	t2=getf(u);
	if(t1!=t2)
	{
		f[t2]=t1;
	}
	return ;
}
int main()
{
	int k,t,x,y,i,j;
	while(scanf("%d %d", &n,&m),n!=0||m!=0)
	{
		init();
		t=0;
		while(m--)
		{
			scanf("%d %d", &k,&x);//先输入数组的大小和数组中的第一个数
			k--;
			while(k--)
			{
				scanf("%d", &y);//依次输入同一数组中的其他元素
				merge(x,y);
			}
		}
		for(i=0; i<n; i++)
		{
			if(getf(i)==getf(0))//要是和0的根节点相同就说明他们在同一个树上
			t++;
		}
		printf("%d\n", t);
	}
	return 0;
}