[备战NOI同步赛]适合ACM-ICPC的并查集模板
并查集模板使用时需要注意一下，不同的题目对并查集的要求不同，某些题目的并查集可能需要携带附加信息，这时需要修改模板的并查集初始化和合并内容，而本并查集模板只支持对并查集结点的父节点、子树深度、子树元素个数进行初始化、合并、查找工作！
/*
并查集模板
By:qpswwww(ZYK)
包含findSet(),makeSet(),Union(),使用路径压缩式查找
注:不同的题目对合并的要求不同，并查集也可能携带附加信息
*/
#define MAXN 30050
int f[MAXN],depth[MAXN],num[MAXN]; //f[i]=点i的父节点,depth[i]=点i的深度,num[i]=i结点对应下面的元素个数
void makeSet(int i) //并查集初始化
{
f[i]=i;
num[i]=1;
depth[i]=1;
}
int findSet(int x) //带路径压缩的并查集查找
{
if(f[x]==x) return x;
return f[x]=findSet(f[x]);
}
void Union(int a,int b) //将a、b合并,有深度优化
{
if(a==b) return;
if(depth[a]>depth[b]) {f[b]=a; num[a]+=num[b];}
else if(depth[a]<depth[b]) {f[a]=b; num[b]+=num[a];}
else {f[a]=b; depth[b]=depth[b]+1; num[b]+=num[a];}
}
例题1:POJ 1611 The Suspects
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.
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.
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
Source
Asia Kaohsiung 2003
题目大意为有多组输入数据，对于每组数据，有n个元素及m组元素，每一组元素对应一个集合，一个元素可存在于多个集合中，与邻元素A在同一集合的所有元素均为A的邻元素，0为最初的邻元素，求每组数据中邻元素的个数
题目可以直接套用上面的目标，下面是代码
/*
并查集模板
By:qpswwww(ZYK)
包含findSet(),makeSet(),Union(),使用路径压缩式查找
注:不同的题目对合并的要求不同，并查集也可能携带附加信息
*/
#include <stdio.h>
#define MAXN 30050
int f[MAXN],depth[MAXN],num[MAXN]; //f[i]=点i的父节点,depth[i]=点i的深度,num[i]=i结点对应下面的元素个数
void makeSet(int i) //并查集初始化
{
f[i]=i;
num[i]=1;
depth[i]=1;
}
int findSet(int x) //带路径压缩的并查集查找
{
if(f[x]==x) return x;
return f[x]=findSet(f[x]);
}
void Union(int a,int b) //将a、b合并,有深度优化
{
if(a==b) return;
if(depth[a]>depth[b]) {f[b]=a; num[a]+=num[b];}
else if(depth[a]<depth[b]) {f[a]=b; num[b]+=num[a];}
else {f[a]=b; depth[b]=depth[b]+1; num[b]+=num[a];}
}
int main()
{
int i,j,k,n,m,a,b,roota,rootb;
while(1)
{
for(i=0;i<MAXN;i++) makeSet(i);
scanf("%d%d",&n,&m);
if(n==0&&m==0) break;
for(i=1;i<=m;i++)
{
scanf("%d%d",&k,&a);
for(j=1;j<k;j++)
{
scanf("%d",&b);
{
roota=findSet(a);
rootb=findSet(b);
Union(roota,rootb); //小组里的人都合并在一个集合里
}
}
}
roota=findSet(0);
printf("%d\n",num[roota]);
}
return 0;
}
例题2:POJ 2524 Ubiquitous Religions
Description
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.
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.
Source
Alberta Collegiate Programming Contest 2003.10.18
题目大意是给出多组数据，对于每组数据，给出n,m,即n个元素,后面接m行,每行有两个数a,b,a和b在同一集合内，求互不相同的集合个数
先将每一对元素合并，所有元素合并完成后，遍历一遍n个元素，找出先根节点的个数，即不同的集合个数
#include <stdio.h>
#define MAXN 50050
int f[MAXN],depth[MAXN],num[MAXN]; //f[i]=点i的父节点,depth[i]=点i的深度,num[i]=i结点对应下面的元素个数
void makeSet(int i) //并查集初始化
{
f[i]=i;
num[i]=1;
depth[i]=1;
}
int findSet(int x) //带路径压缩的并查集查找
{
if(f[x]==x) return x;
return f[x]=findSet(f[x]);
}
void Union(int a,int b) //将a、b合并,有深度优化
{
if(a==b) return;
if(depth[a]>depth[b]) {f[b]=a; num[a]+=num[b];}
else if(depth[a]<depth[b]) {f[a]=b; num[b]+=num[a];}
else {f[a]=b; depth[b]=depth[b]+1; num[b]+=num[a];}
}
int main()
{
int i,j,n,m,a,b,tot=0,sum,roota,rootb;
while(++tot)
{
sum=0;
for(i=0;i<MAXN;i++) makeSet(i);
scanf("%d%d",&n,&m);
if(n==0&&m==0) break;
for(i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
roota=findSet(a);
rootb=findSet(b);
Union(roota,rootb);
}
for(i=1;i<=n;i++)
if(findSet(i)==i) sum++; //统计不同集合(根节点)个数
printf("Case %d: %d\n",tot,sum);
}
return 0;
}