HDOJ 1213 How Many Tables（并查集）

How Many Tables

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

题目大意：有N个人一起吃饭，只有互相认识的人才能坐在同一桌。例如A认识B，B认识C，那么A、B、C就可以坐在一桌；如果A认识B，B认识C，D认识E，那么A、B、C一桌，D、E一桌。给出人数与他们之间的关系，求至少需要几张桌子。

解题思路：并查集的基础应用。将互相认识的人放在同一个集合里，然后统计父节点为自己的节点数即可。

代码如下：

#include <algorithm>#include <cctype>#include <climits>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <queue>#include <set>#include <stack>#include <vector>#define EPS 1e-6#define INF INT_MAX / 10#define LL long long#define MOD 100000000#define PI acos(-1.0)const int maxn = 1005;int par[maxn],rank[maxn];void init(int n){    for(int i = 1;i <= n;i++){        par[i] = i;        rank[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(rank[x] < rank[y]){        par[x] = y;    }    else{        par[y] = x;        if(rank[x] == rank[y]){            rank[x]++;        }    }}int main(){    int t,n,m,x,y;    scanf("%d",&t);    while(t--){        scanf("%d %d",&n,&m);        init(n);        while(m--){            scanf("%d %d",&x,&y);            unite(x,y);        }        int ans = 0;        for(int i = 1;i <= n;i++){            if(par[i] == i)                ans++;        }        printf("%d\n",ans);    }}