Solution Notes: This problem is solved via recursive depth-first search: Our current state is described by 4 numbers, giving the (x,y) offset of objects 2 and 3 relative to their initial positions. Each move we make, we can change one of these numbers by +1 or -1 to slide either object 2 or 3 (as long as this doesn't create any overlap between the objects), or we can change both x offsets or both y offsets by +1 or -1 to simulate moving object 1 (since moving object 1 is equivalent to keeping object 1 fixed and moving both objects 2 and 3 in the opposite direction).

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <set>
#include <algorithm>
using namespace std;

int n[3];
int lis[3][100][2];

int ok[3][3][51][51];


int dir[4][2] = {-1, 0, 1, 0, 0, 1, 0, -1};

char seen[42][42][42][42];

set<pair<int, int> > occupied;
int check(int i1, int i2, int dx, int dy) {
	occupied.clear();
	for(int i = 0; i < n[i1]; ++i) {
		occupied.insert(make_pair(lis[i1][i][0], lis[i1][i][1]));
	}
	for(int i = 0; i < n[i2]; ++i) {
		if(occupied.find(make_pair(lis[i2][i][0] + dx, lis[i2][i][1] + dy)) != occupied.end()) {
			return 0;
		}
	}
	return 1;
}

int isOkPair(int i1, int i2, int dx, int dy) {
	if(dx < -20 || dx > 20 || dy < -20 || dy > 20) return 1;
	return ok[i1][i2][dx + 20][dy + 20];
}

int isOk(int dx1, int dy1, int dx2, int dy2) {
	return isOkPair(0, 1, dx1, dy1) &&
			isOkPair(0, 2, dx2, dy2) &&
			isOkPair(1, 2, dx2 - dx1, dy2 - dy1);	
}

const int LIM = 20;

char q[3000000][4];
int tail;

void add (int dx1, int dy1, int dx2, int dy2) {
	if(dx1 < -LIM || dy1 < -LIM || dx1 > LIM || dy1 > LIM ||
	   dx2 < -LIM || dy2 < -LIM || dx2 > LIM || dy2 > LIM) return;
	if(!isOk(dx1, dy1, dx2, dy2)) return;
	if(seen[dx1 + LIM][dy1 + LIM][dx2 + LIM][dy2 + LIM]) return;
	seen[dx1 + LIM][dy1 + LIM][dx2 + LIM][dy2 + LIM] = 1;
	q[tail][0] = dx1;
	q[tail][1] = dy1;
	q[tail][2] = dx2;
	q[tail][3] = dy2;
	tail++;
}

int main() {
        freopen("unlock.in", "r", stdin);
        freopen("unlock.out", "w", stdout);
	scanf("%d%d%d", &n[0], &n[1], &n[2]);	
	for(int i = 0; i < 3; ++i) {
		for(int j = 0; j < n[i]; ++j) {
			scanf("%d%d", &lis[i][j][0], &lis[i][j][1]);
		}
	}
	for(int i1 = 0; i1 < 3; ++i1) 
	for(int i2 = 0; i2 < i1; ++i2) 
		for(int dx = -20; dx <= 20; ++dx)
		for(int dy = -20; dy <= 20; ++dy) {
			ok[i2][i1][dx + 20][dy + 20] = check(i1, i2, dx, dy);
		}
	memset(seen, 0, sizeof(seen));
	tail = 0;
	add(0, 0, 0, 0);
	for(int i = 0; i < tail; ++i) {
		int dx1 = q[i][0];
		int dy1 = q[i][1];
		int dx2 = q[i][2];
		int dy2 = q[i][3];
		if(dx1 == LIM && dy1 == LIM && dx2 == -LIM && dy2 == -LIM) {
			puts("1");
			return 0;
		}
		for(int j = 0; j < 4; ++j) {
			add(dx1 + dir[j][0], dy1 + dir[j][1], dx2, dy2);
			add(dx1, dy1, dx2 + dir[j][0], dy2 + dir[j][1]);
			add(dx1 + dir[j][0], dy1 + dir[j][1],
				dx2 + dir[j][0], dy2 + dir[j][1]);
		}
	}
	puts("0");
    return 0;
}