题解 CF995A 【Tesla】

考虑把刚好在车位前面的车停进去，然后剩下的车统一轮换，能停的时候就停进去，不断重复这个过程。

因为中间的车道是 $2\times n$ 的格子，最坏情况一辆车要转一圈才能找到车位，找到车位后还要一次操作停进去，总共 $k$ 辆车，最坏的操作次数为 $k\times(2\times n+1)\le 10100\lt 20000$，因此这种方法可行。

无解的话就判断没有车被轮换的时候是全都停到了车位里还是全都动不了即可。

//By: Luogu@rui_er(122461)
#include <bits/stdc++.h>
#define rep(x,y,z) for(int x=y;x<=z;x++)
#define per(x,y,z) for(int x=y;x>=z;x--)
#define debug printf("Running %s on line %d...\n",__FUNCTION__,__LINE__)
using namespace std;
typedef long long ll;
const int N = 205; 

int n, m, a[5][N];
tuple<int, int> s[N], t[N];
vector<tuple<int, int, int> > ans;
template<typename T> void chkmin(T &x, T y) {if(x > y) x = y;}
template<typename T> void chkmax(T &x, T y) {if(x < y) x = y;}
tuple<int, int> getNext(const tuple<int, int>& pos) {
	int x = get<0>(pos), y = get<1>(pos);
	if(x == 2) {
		if(y == n) return make_tuple(3, n);
		return make_tuple(2, y+1);
	}
	if(y == 1) return make_tuple(2, 1);
	return make_tuple(3, y-1);
}
bool isEmpty(const tuple<int, int>& pos) {
	return !a[get<0>(pos)][get<1>(pos)];
}
bool checkPark(const int& id) {
	if(abs(get<0>(s[id])-get<0>(t[id])) == 1 && get<1>(s[id]) == get<1>(t[id])) {
		ans.push_back(make_tuple(id, get<0>(t[id]), get<1>(t[id])));
		a[get<0>(s[id])][get<1>(s[id])] = a[get<0>(t[id])][get<1>(t[id])] = 0;
		return 1;
	}
	return 0;
}
bool rotateR() {
	bool ok = 0;
	rep(i, 2, 3) {
		rep(j, 1, n) {
			if(!a[i][j]) continue;
			tuple<int, int> nxt = getNext(make_tuple(i, j));
			if(checkPark(a[i][j])) ok = 1;
			else if(isEmpty(nxt)) {
				int id = a[i][j];
				ans.push_back(make_tuple(id, get<0>(nxt), get<1>(nxt)));
				a[i][j] = 0;
				a[get<0>(nxt)][get<1>(nxt)] = id;
				s[id] = nxt;
				ok = 1;
			}
		}
	}
	return ok;
}

int main() {
	scanf("%d%d", &n, &m);
	rep(i, 1, 4) rep(j, 1, n) {
		scanf("%d", &a[i][j]);
		if(a[i][j]) {
			if(i == 1 || i == 4) t[a[i][j]] = make_tuple(i, j);
			else s[a[i][j]] = make_tuple(i, j);
		}
	}
	while(rotateR())
		;
	bool tag = 1;
	rep(i, 1, 4) {
		rep(j, 1, n) if(a[i][j]) {tag = 0; break;}
		if(!tag) break;
	}
	if(!tag) puts("-1");
	else {
		int sz = ans.size();
		printf("%d\n", sz);
		for(auto p : ans) printf("%d %d %d\n", get<0>(p), get<1>(p), get<2>(p));
	}
	return 0;
}