正文
杭电 1213 How Many Tables （并查集求团体数）
小程序：扫一扫查出行【扫一扫了解最新限行尾号】
复制小程序
DescriptionToday is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
InputThe input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.
OutputFor each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
Sample Input2
5 3
1 2
2 3
4 55 1
2 5
2
5 3
1 2
2 3
4 55 1
2 5
Sample Output2
4大意：
有n个人，m中关系，每种关系的人在一起，问有多少种团体；第一组测试数据有5个人编号1~5，3种关系1和2,2和3,4和5，所以1,2,3为一个团体，4,5为一个团体。将有关系的人放在一个集合，求出集合数。
并查集模板
#include<cstdio>
int n,m,ans,i,a,b,fa[]; /*int find(int a)
{
int r=a;
while(r != fa[r])
{
r=fa[r];
}
int i=a,j; //循环实现路径压缩
while(i != r)
{
j=fa[i];
fa[i]=r;
i=j;
}
return r;
}*/ int find(int a) //递归实现路径压缩
{
if(a == fa[a])
{
return a;
}
else
{
return fa[a]=find(fa[a]);
}
} void f1(int x,int y)
{
int nx,ny;
nx=find(x);
ny=find(y);
if(nx != ny)
{
fa[ny] = nx;
}
} int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&n,&m);
for(i = ; i <= n ; i++)
{
fa[i] = i;
}
for(i = ; i < m ; i++)
{
scanf("%d %d",&a,&b);
f1(a,b);
}
ans=;
for(i = ; i <= n ; i++)
{
if(fa[i] == i)
{
ans++;
}
}
printf("%d\n",ans);
}
}
2
4大意：
有n个人，m中关系，每种关系的人在一起，问有多少种团体；第一组测试数据有5个人编号1~5，3种关系1和2,2和3,4和5，所以1,2,3为一个团体，4,5为一个团体。将有关系的人放在一个集合，求出集合数。
并查集模板
#include<cstdio>
int n,m,ans,i,a,b,fa[]; /*int find(int a)
{
int r=a;
while(r != fa[r])
{
r=fa[r];
}
int i=a,j; //循环实现路径压缩
while(i != r)
{
j=fa[i];
fa[i]=r;
i=j;
}
return r;
}*/ int find(int a) //递归实现路径压缩
{
if(a == fa[a])
{
return a;
}
else
{
return fa[a]=find(fa[a]);
}
} void f1(int x,int y)
{
int nx,ny;
nx=find(x);
ny=find(y);
if(nx != ny)
{
fa[ny] = nx;
}
} int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&n,&m);
for(i = ; i <= n ; i++)
{
fa[i] = i;
}
for(i = ; i < m ; i++)
{
scanf("%d %d",&a,&b);
f1(a,b);
}
ans=;
for(i = ; i <= n ; i++)
{
if(fa[i] == i)
{
ans++;
}
}
printf("%d\n",ans);
}
}
