HDU 1213(并查集)

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27812    Accepted Submission(s): 13801

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

 

Author

Ignatius.L

 

Source

杭电ACM省赛集训队选拔赛之热身赛

 

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/*
并查集入门题，无坑点。
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = ;
int fa[maxn];
int T, N, M;
int u, v;
int Find(int u)
{
    if(u == fa[u]) return fa[u];
    else return fa[u] = Find(fa[u]);
}
void adde(int u, int v)
{
    int x = Find(u);
    int y = Find(v);
    if(x != y) fa[x] = fa[y];
}
int main ()
{
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &N, &M);
        for(int i = ; i <= N; i++)
            fa[i] = i;
        for(int i = ; i <= M; i++)
        {
            scanf("%d%d", &u, &v);
            adde(u, v);
        }
        int ans = ;
        for(int i = ; i <= N; i++)
        {
            if(fa[i] == i) ans++;
        }
        printf("%d\n", ans);
    }
    return ;
}