HDU 1213

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1213

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

并查集的模板题。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000+5;

int n,m;

int par[maxn],ran[maxn];
void init(int l,int r){for(int i=l;i<=r;i++) par[i]=i,ran[i]=0;}
int find(int x){return (par[x]==x)?x:(par[x]=find(par[x]));}
void unite(int x,int y)
{
    x=find(x), y=find(y);
    if(x==y) return;
    if(ran[x]<ran[y]) par[x]=y;
    else par[y]=x, ran[x]+=(ran[x]==ran[y]);
}
bool isSame(int x,int y){return find(x)==find(y);}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);

        init(1,n);
        for(int i=1,a,b;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            if(!isSame(a,b)) unite(a,b);
        }

        set<int> S;
        for(int i=1;i<=n;i++) S.insert(find(i));

        printf("%d
",S.size());
    }
}

整理一下并查集的两种模板吧：

int par[maxn],ran[maxn];
void init(int l,int r){for(int i=l;i<=r;i++) par[i]=i,ran[i]=0;}
int find(int x){return (par[x]==x)?x:(par[x]=find(par[x]));}
void unite(int x,int y)
{
    x=find(x), y=find(y);
    if(x==y) return;
    if(ran[x]<ran[y]) par[x]=y;
    else par[y]=x, ran[x]+=(ran[x]==ran[y]);
}
bool isSame(int x,int y){return find(x)==find(y);}

int par[maxn];
void init(int l,int r){for(int i=l;i<=r;i++) par[i]=i;}
int find(int x){return (par[x]==x)?x:(par[x]=find(par[x]));}

//这种简单的并查集的合并方式是:
int t1=find(u),t2=find(v);
if(t1!=t2) par[t1]=t2; //这种是把点u所在树并入点v所在树
if(t1!=t2) par[t2]=t1; //这种是把点v所在树并入点u所在树