noip模拟赛 轮换

分析：模拟题，关键就是要理解题目意思.m≥3的轮换可以拆成m=2的小轮换，小轮换的话只需要交换一下就可以了.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int n, p, k, f[][], a[], pos[];

int main()
{
    scanf("%d%d%d", &n, &p, &k);
    for (int i = ; i <= n; i++)
        a[i] = i, pos[i] = i;
    for (int i = p; i >= ; i--)
    {
        scanf("%d", &f[i][]);
        for (int j = ; j <= f[i][]; j++)
            scanf("%d", &f[i][j]);
    }
    for (int i = ; i <= p; i++)
        for (int j = ; j <= f[i][]; j++)
        {
            swap(a[pos[f[i][]]], a[pos[f[i][j]]]);
            swap(pos[f[i][]], pos[f[i][j]]);
        }
    for (int i = ; i <= n; i++)
        printf("%d ", a[i]);

    return ;
}