How Many Tables
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
并查集:find和union
构建一个含有n棵树的森林。用简单的数组实现。对于每次查找看两颗树的能否进行合并(找他们的根是否相等)
#include<iostream>
using namespace std;
int pre[1005];
int sum;
int finder(int x)
{
if(pre[x]==x) return x;
else pre[x]=finder(pre[x]);
}
void unioner(int a,int b)
{
int s=finder(a);
int t=finder(b);
if(s==t) return;
sum-=1;
pre[s]=t;
}
int main()
{
int n,m,i,cnt;
int c,d;
cin>>cnt;
while(cnt--)
{
cin>>n>>m;
for(i=1;i<=n;i++)
{
pre[i]=i;
}
sum=n;
for(i=0;i<m;i++)
{
cin>>c>>d;
unioner(c,d);
}
cout<<sum<<endl;
}
return 0;
}
using namespace std;
int pre[1005];
int sum;
int finder(int x)
{
if(pre[x]==x) return x;
else pre[x]=finder(pre[x]);
}
void unioner(int a,int b)
{
int s=finder(a);
int t=finder(b);
if(s==t) return;
sum-=1;
pre[s]=t;
}
int main()
{
int n,m,i,cnt;
int c,d;
cin>>cnt;
while(cnt--)
{
cin>>n>>m;
for(i=1;i<=n;i++)
{
pre[i]=i;
}
sum=n;
for(i=0;i<m;i++)
{
cin>>c>>d;
unioner(c,d);
}
cout<<sum<<endl;
}
return 0;
}
为了明天所以选择坚定的执着今天。