How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31632 Accepted Submission(s): 15706
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.
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
解题心得
- 题意就是给你一些关系，求有多少个相互之间没有关系的集合。
- 这是一个很简单的并查集的应用，直接将该合并的合并，然后直接找根，有多少个不同的根就有需要多少个桌子。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1010;
int father[maxn];
int n,m;
int find(int x)
{
if(father[x] == x)
return x;
else
return father[x] = find(father[x]);
}
void merge(int a,int b)
{
int fa = find(a);
int fb = find(b);
if(fa != fb)
father[fa] = fb;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int ans = 0;
for(int i=0;i<maxn;i++)
father[i] = i;
scanf("%d%d",&n,&m);
while(m--)
{
int a,b;
scanf("%d%d",&a,&b);
if(find(a) != find(b))//两个之间是朋友则直接合并
{
merge(a,b);
}
}
for(int i=1;i<=n;i++)
if(father[i] == i)//在n个人中有多少个没有联系的集合就是求有多少个根
ans ++;
printf("%d
",ans);
}
}