CSUOJ 1208-排队

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1208

亏我还想用并查集去做，这道题用数组模拟链表就行，数组下标为编号，一个一个接下去就行，

在接的同时计数。这题有点恶心的是必须把数组开到101000以上，要不就会RE，这也是考查

的一个方面吧！

#include<iostream>
#include<string.h>
using namespace std;
const int N = 101005;
int next[N];

int main()
{
    int n, m;
    int a, b;
    while( cin >> n >> m)
    {
        memset( next, 0, sizeof(next) );
        for( int i = 1; i <= m; i ++)
        {
            cin >> a >> b;
            next[a] = b;
        }
        int min = 1000001, k;
        for( int i = 1; i <= n; i ++)
        {
            int cnt = 0;
            for( int j = next[i]; j; j = next[j])
                cnt ++;
            if( cnt < min) {
                min = cnt;
                k = i;
            }
        }
        cout << k << endl;
    }
    return 0;
}

 下面是并查集：

#include<stdio.h>
#define N 101005

int p[N], cnt[N], n, m;

int find_set( int x)
{
    return p[x] == x ? x : ( p[x] = find_set( p[x]) );
}

void union_set( int x, int y)
{
    x = find_set( x);
    y = find_set( y);
    if( x == y) return;
    else if( x > y) {
        p[x] = y;
        cnt[y] += cnt[x];
    }
    else {
        p[y] = x;
        cnt[x] += cnt[y];
    }
}

int main()
{
    int a, b;
    while( scanf( "%d%d", &n, &m) == 2)
    {
        int i;
        for( i = 1; i <= n; i ++)
        {
            p[i] = i;
            cnt[i] = 0;
        }
        for( ; i <= n + m; i ++)
        {
            p[i] = i;
            cnt[i] = 1;
        }
        while( m --)
        {
            scanf( "%d%d", &a, &b);
            union_set( a, b);
        }

        int min = cnt[1], k = 1;
        for( i = 2; i <= n; i ++)
        {
            if( cnt[i] < min) {
                min = cnt[i];
                k = i;
            }
        }
        printf( "%d\n", k);
    }
    return 0;
}