How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16616 Accepted Submission(s):
8137
Problem Description
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.
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.
Input
The 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.
Output
For each test case, just output how many tables
Ignatius needs at least. Do NOT print any blanks.
Sample Input
2
5 3
1 2
2 3
4 5
5 1
2 5
Sample Output
2
4
题解同hdu1232
#include<stdio.h>
int set[2100];
int find(int fa) //找到根节点
{
int ch=fa;
int t;
while(fa!=set[fa])
fa=set[fa];
while(ch!=fa)
{
t=set[ch];
set[ch]=fa;
ch=t;
}
return fa;
}
int mix(int x,int y) //合并已有人数
{
int fx,fy;
fx=find(x);
fy=find(y);
if(fx!=fy)
set[fx]=fy;
}
int main()
{
int n,m,j,i,table,people,s;
scanf("%d",&n);
while(n--)
{
scanf("%d %d",&people,&table);
for(i=1;i<=people;i++)
set[i]=i;
for(i=1;i<=table;i++)
{
scanf("%d %d",&m,&j);
mix(m,j);
}
s=0;
for(i=1;i<=people;i++)
{
if(set[i]==i)
s++; //需要的桌子数
}
printf("%d
",s);
}
return 0;
}