【題解】ZeroJudge d831: 畢業旅行

#include <iostream>
using namespace std;
int p[1000005];
int n, m, mx, a, b;

int find(int a){
    if (p[a] < 0) return a;
    else {
        p[a] = find(p[a]);
        return p[a];
    }
}

int main() {
    while (cin >> n >> m){
        for (int i = 0; i < n; i++){
            p[i] = -1;
        }
        mx = 1;
        for (int i = 0; i < m; i++){
            cin >> a >> b;
            a = find(a);
            b = find(b);
            if (a != b){
                p[a] += p[b];
                mx = max(mx, -p[a]);
                p[b] = a;
            }
        }
        cout << mx << "\n";
    }
}

def Find(x):
    # 初始值 p[x] = -1 表示指向自己
    if p[x] < 0:
        return x
    else:
        # 順便做「路徑壓縮」
        p[x] = Find(p[x])
        return p[x]

def Union(x, y):
    g1 = Find(x)
    g2 = Find(y)
    if g1 != g2:
        # 加總元素個數(負值)
        # p[]:只有「代表元素」的值是存「元素個數」，其它的存「代表元素編號」
        p[g1] += p[g2]
        p[g2] = g1
#=== main ===
while True:
    try:      
        N, M = map(int, input().split())
        # 初始值 p[x] = -1 表示自成一群，「代表元素」是自己
        p = [-1 for i in range(N)]
        mx = 1
        for _ in range(M):
            a, b = map(int, input().split())
            g1 = Find(a)
            g2 = Find(b)
            if (g1 != g2):
                p[g1] += p[g2]
                p[g2] = g1
                mx = max(mx, -p[g1])
        print(mx)
    except:
        break