hdu1213(基础并查集)

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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

25 31 22 34 55 12 5

Sample Output

24
题意：有n个人，m个关系，相互认识的人可以坐在一张桌子上，问你需要多少张桌子（相互认识：假如A和B是盆友，B和C是盆友，那么A和C是盆友）。
思路：基础并查集，相互认识的人分在一个集合里面，最后统计有多少个集合即可。
代码：
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
25 31 22 34 55 12 5
 

Sample Output
24

#include<bits/stdc++.h>using namespace std;const int maxn=1005;int father[maxn];int findset(int x)//这里有时候递归会爆栈要注意{    if(father[x]!=x)        father[x]=findset(father[x]);    return father[x];}void Union(int a,int b){    int fa=findset(a);    int fb=findset(b);    if(fa==fb) return;//属于同一个集合    father[fa]=fb;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,m;        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++) father[i]=i;//初始化并查集        for(int i=0;i<m;i++)        {            int x,y;            scanf("%d%d",&x,&y);            Union(x,y);        }        int ans=0;        for(int i=1;i<=n;i++)        {            if(father[i]==i) ans++;//属于不同集合        }        printf("%d\n",ans);    }    return 0;}