【HD 1213】How Many Tables

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24377 Accepted Submission(s): 12202

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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和畅通工程完全一样的题，本质上是求根节点数，有多少个根节点就说明有多少桌。详解可以查看 -> 传送门<-对，就是他

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#include<cstdio>#include<cstring>//一样求根节点数目using namespace std;const int MAX_N = 1005; int par[MAX_N];int rank[MAX_N];void init( int n ){    for( int i=1; i<=n; i++ )    {        par[i] = i;        rank[i] = 0;    }} int find( int x ){    if( par[x] == x )    return x;    else    return find(par[x]);}void unite( int x,int y ){    x = find(x);    y = find(y);    if( x == y ) return;    if( rank[x] < rank[y] )    {        par[x] = y;    }else{        par[y] = x;        if( rank[x] == rank[y] ) rank[x]++;    }//  par[x]=y;}int main(){    int n,m,a,b;    int t;     scanf("%d",&t);    while( t-- )    {        scanf("%d",&n);         init(n);        scanf("%d",&m);        for( int i=0; i<m; i++ )        {            scanf("%d%d",&a,&b);            unite(a,b);        }        int count=0;        for( int i=1; i<=n; i++ )        {            if( par[i] == i )            count++;        }        printf("%d\n",count);    }}