HDU 1213 How Many Tables [并查集]

Posted 2021-10-02 布图

题目

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

解释与代码

这里我用了set容器来存,最后遍历一遍即可

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <iostream>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <bitset>
#include <vector>
#include <limits.h>
#include <assert.h>
#include <functional>
#include <numeric>
#include <ctime>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define pb          push_back
#define ppb         pop_back
#define lbnd        lower_bound
#define ubnd        upper_bound
#define endl        '\\n'
#define trav(a, x)  for(auto& a : x)
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define DEBUG       cerr<<"/n>>>I'm Here<<</n"<<endl;
#define display(x)  trav(a,x) cout<<a<<" ";cout<<endl;
#define what_is(x)  cerr << #x << " is " << x << endl;
#define ini(a)      memset(a,0,sizeof(a))
#define ini2(a,b)   memset(a,b,sizeof(a))
#define case        ll T;read(T);for(ll Q=1;Q<=T;Q++)
#define lowbit(x)   x&(-x)
#define pr          printf
#define sc          scanf
#define _           0
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define DBG(x) \\
    (void)(cout << "L" << __LINE__ \\
    << ": " << #x << " = " << (x) << '\\n')
#define TIE \\
    cin.tie(0);cout.tie(0);\\
    ios::sync_with_stdio(false);
//#define long long int

//using namespace __gnu_pbds;

template <typename T>  
void read(T &x) {  
    x = 0;  
    int f = 1;  
    char ch = getchar();  
    while (!isdigit(ch)) {  
        if (ch == '-') f = -1;  
        ch = getchar();  
    }  
    while (isdigit(ch)) { 
        x = x * 10 + (ch ^ 48);  
        ch = getchar();  
    }  
    x *= f;  
    return;  
}

inline void write(long long x) {
    if(x<0) putchar('-'), x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
    putchar('\\n');
}

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI    = acos(-1.0);
const double eps   = 1e-6;
const int    INF   = 0x3f3f3f3f;
const ll     LLINF = 0x3f3f3f3f3f3f3f3f;
const int    maxn  = 1099;
const ll     N     = 5;

int fa[maxn];

set<int> se;

int get(int x) {
	if (x == fa[x]) return x;
	return fa[x] = get(fa[x]);
}

void solve(){
	se.clear();
	int n, m;
	cin>>n>>m;
	for (int i=0; i<maxn; i++) fa[i] = i;
	for (int i=0; i<m; i++) {
		int x, y;
		cin>>x>>y;
		x = get(x);
		y = get(y);
		if (x != y) {
			fa[x] = y;
		}
	}
	for (int i=1; i<=n; i++) {
		int x = get(i);
		if (se.count(x) == 0) {
			se.insert(x);
		}
	}
	cout<<se.size()<<endl;
}


int main()
{
//	TIE;
    #ifndef ONLINE_JUDGE
//    freopen ("in.txt" , "r", stdin );
//    freopen ("out.txt", "w", stdout);
    #else
    #endif
//	solve();
    case{solve();}
//    case{cout<<"Case "<<Q<<":"<<endl;solve();}
	return ~~(0^_^0);
}