并查集-D-宗教何其多

There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.

You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.
Input
The input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.
Output
For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.
Sample Input
10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0
Sample Output
Case 1: 1
Case 2: 7
Hint
Huge input, scanf is recommended.

#include<stdio.h>
int a[50009];

int find(int x){
	int p=x;
	while(a[p]!=p){
		p=a[p];
	}
	int i=x,j;
	if(i!=p){
		j=a[i];
		a[i]=p;
		i=j;
	}
	return p;
}

void judge(int x,int y){
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy){
		a[fx]=fy;
	}
}


int main(){
	int n,m;
	int i,j=1;
	int x,y;
	while(scanf("%d %d",&n,&m)!=EOF&&n!=0&&m!=0){
		int count=0;
		for(i=1;i<=n;i++){
			a[i]=i;
		}
		while(m--){
			scanf("%d %d",&x,&y);
			judge(x,y);
		}
		for(i=1;i<=n;i++){
			if(a[i]==i){
				count++;
			}
		}
		printf("Case %d: %d\n",j,count);
		j++;
	}
	return 0;
}
在Java Servlet开发中遇到'javax.naming.NamingException: Cannot create resource instance'异常时,可以通过以下步骤进行诊断和解决问题: 参考资源链接:[Apache命名解析异常:创建资源实例失败](https://wenku.youkuaiyun.com/doc/5akqpm2ysp?spm=1055.2569.3001.10343) 1. 首先检查应用服务器的配置文件,如web.xml、context.xml或application.properties,确保JNDI资源的配置信息准确无误,包括JNDI名称、资源工厂类以及资源URL等。 2. 查看服务器日志文件,特别关注与JNDI相关的错误信息和堆栈跟踪,这些信息将帮助你定位问题的具体位置。 3. 验证依赖的第三方库是否与当前应用兼容,确保没有版本冲突,并更新至最新版本以解决潜在的兼容性问题。 4. 确认应用程序有足够权限进行资源创建,检查服务器配置文件中的权限设置,如user、role以及相关的access权限。 5. 审查涉及JNDI操作的代码,特别是资源工厂类中的`getObjectInstance()`方法,检查是否有逻辑错误或配置错误。 6. 如果怀疑是服务器初始化或部署问题,尝试重启服务,有时候重启可以解决配置或环境初始化的问题。 7. 如果问题仍然存在,可以考虑创建最小化复现案例,并在社区或专业论坛中寻求帮助。 建议阅读《Apache命名解析异常:创建资源实例失败》一书,该资料详细介绍了 NamingException 异常的背景、原因以及解决方案,对开发人员在解决此类问题时提供了系统性的指导和帮助。 参考资源链接:[Apache命名解析异常:创建资源实例失败](https://wenku.youkuaiyun.com/doc/5akqpm2ysp?spm=1055.2569.3001.10343)
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