【hdu 1213 基础并查集】How Many Tables

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29112    Accepted Submission(s): 14436

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

ac代码：

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;int pre[1010];bool gg[1010];int findd(int x)                           //查找根节点{    int r=x;    while ( pre[r ] != r )                //返回根节点 r        r=pre[r ];    int i=x, j ;    while( i != r )                       //路径压缩    {        j = pre[ i ]; // 在改变上级之前用临时变量j记录下他的值        pre[ i ]= r ; //把上级改为根节点        i=j;    }    return r ;}void join(int x,int y)                     //判断x y是否连通，//如果已经连通，就不用管了 //如果不连通，就把它们所在的连通分支合并起,{    int fx=findd(x),fy=findd(y);    if(fx!=fy)        pre[fx ]=fy;}int main(){    int t;    cin>>t;    while(t--)    {        int n,m;        cin>>n>>m;        for(int i=1; i<=n; i++)      //初始化        {            pre[i]=i;        }        int a,b,ans=0;        for(int i=0; i<m; i++)       // 吸收数据        {            cin>>a>>b;            join(a,b);        }        memset(gg,0,sizeof(gg));        for(int i=1; i<=n; i++)        {            int x=findd(i);         //标记根节点            gg[x]=1;        }        for(int i=1; i<=n; i++)        {            if(gg[i]==1) ans++;        }        cout<<ans<<endl;    }    return 0;}